Question 4015 Marks
If $\text{y}=(\text{x}-1)\log(\text{x}-1)-(\text{x}+1)\log(\text{x}+1)$ prove that $\frac{\text{dy}}{\text{dx}}=\log\Big(\frac{\text{x}-1}{1+\text{x}}\Big)$
AnswerWe have, $\text{y}=(\text{x}-1)\log(\text{x}-1)-(\text{x}+1)\log(\text{x}+1)$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[(\text{x}-1)\log(\text{x}-1)-(\text{x}+1)\log(\text{x}+1)\big]$
$=\Big[(\text{x}-1)\frac{\text{d}}{\text{dx}}\log(\text{x}-1)+\log(\text{x}-1)\frac{\text{d}}{\text{dx}}(\text{x}-1)\Big] \\ -\Big[(\text{x}+1)\frac{\text{d}}{\text{dx}}\log(\text{x}+1)+\log(\text{x}+1)\frac{\text{d}}{\text{dx}}(\text{x}+1)\Big]$
$=\Big[(\text{x}-1)\times\frac{1}{(\text{x}-1)}\frac{\text{d}}{\text{dx}}(\text{x}-1)+\log(\text{x}-1)\times(1)\Big] \\ -\Big[(\text{x}+1)\times\frac{1}{(\text{x}+1)}\times\frac{\text{d}}{\text{dx}}(\text{x}+1)+\log(\text{x}+1)(1)\Big]$
$=\big[1+\log(\text{x}-1)\big]-\big[1+\log(\text{x}+1)\big]$
$=\log(\text{x}-1)-\log(\text{x}+1)$
$=\log\frac{(\text{x}-1)}{(\text{x}+1)}$
So,
$\frac{\text{dy}}{\text{dx}}=\log\frac{(\text{x}-1)}{(\text{x}+1)}$
View full question & answer→Question 4025 Marks
Differentiate the following functions with respect to x:
$(\log\text{x})^{\log\text{x}}$
AnswerLet $\text{y}=(\log\text{x})^{\log\text{x}}\ .....(\text{i})$
Taking logarithm on both the sides, we obtain
$\log\text{y}=\log\text{x}.\log(\log\text{x})$
Differentiating both sides with resepect to x, we obtain
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\log\text{x}.\log(\log\text{x})\big]$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log(\log\text{x}).\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}.\frac{\text{d}}{\text{dx}}[\log(\log\text{x})]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\log(\log\text{x}).\frac{1}{\text{x}}+\log\text{x}.\frac{1}{\log\text{x}}.\frac{\text{d}}{\text{dx}}(\log\text{x})\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{1}{\text{x}}\log(\log\text{x})+\frac{1}{\text{x}}\Big]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=(\log\text{x})^{\log\text{x}}\Big[\frac{1}{\text{x}}+\frac{\log(\log\text{x})}{\text{x}}\Big]$
View full question & answer→Question 4035 Marks
If $\text{x}=\text{a}\sin\text{t}\ \text{and}\ \text{y}=\text{a}(\cos\text{t}+\log\tan\frac{\text{t}}{2}),$ find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
Answer$\text{x}=\text{a}\sin\text{t}\ \text{and}\ \text{y}=\text{a}(\cos\text{t}+\log\tan\frac{\text{t}}{2}),$
$\frac{\text{dx}}{\text{dt}}=\text{a}\cos\text{t}$
$\frac{\text{d}^2\text{x}}{\text{dt}^2}=-\text{a}\sin\text{t}$
$\frac{\text{dy}}{\text{dt}}=-\text{a}\sin\text{t}+\text{a}\frac{1}{\tan\frac{\text{t}}{2}}\times\sec^2\frac{\text{t}}{2}\times\frac{1}{2}$
$=-\text{a}\sin\text{t}+\text{a}\frac{1}{2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}$
$=-\text{a}\sin\text{t}+\text{a}\ \text{cosec}\ \text{t}$
$\frac{\text{d}^2\text{y}}{\text{dt}^2}=-\text{a}\cos\text{t}-\text{a cosec t}\cot\text{t}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\frac{\text{dx}}{\text{dt}}\frac{\text{d}^2\text{y}}{\text{dt}^2}-\frac{\text{dy}}{\text{dt}}\frac{\text{d}^2\text{x}}{\text{dt}^2}}{\Big(\frac{\text{dx}}{\text{dt}}\Big)^3}$
$=\frac{\text{a}\cos\text{t}(-\text{a}\cos\text{t}-\text{a}\ \text{cosec t}\cot\text{t})-(-\text{a}\sin\text{t}+\text{a}\text{cosec t})(-\text{a}\sin\text{t})}{(\text{a}\cos\text{t})^3}$
$=\frac{-\text{a}^2\cos^2\text{t}-\text{a}^2\cot^2\text{t}-\text{a}^2\sin^2\text{t}+\text{a}^2}{\text{a}^3\cos^3\text{t}}$
$=\frac{-\text{a}^2\cos^2\text{t}-\text{a}^2\sin^2\text{t}-\text{a}^2\cot^2\text{t}+\text{a}^2}{\text{a}^3\cos^3\text{t}}$
$=\frac{-\text{a}^2(\cos^2\text{t}+\sin^2\text{t})-\text{a}^2\cot^2\text{t}+\text{a}^2}{\text{a}^3\cos^3\text{t}}$
$=-\frac{1}{\text{a}\sin^2\text{t}\cos\text{t}}$
View full question & answer→Question 4045 Marks
If $(\sin\text{x})^{\text{y}}=(\cos\text{y})^{\text{x}},$ Prove that $\frac{\text{dy}}{\text{dx}}=\frac{\log\cos\text{y}-\text{y}\cot\text{x}}{\log\sin\text{x}+\text{x}\tan\text{y}}$
AnswerWe have, $(\sin\text{x})^{\text{y}}=(\cos\text{y})^{\text{x}}$
Taking log on both sides,
$\log(\sin\text{x})^\text{y}=\log(\cos\text{y})^{\text{x}}$
$\Rightarrow\text{y}\log(\sin\text{x})=\text{x}\log(\cos\text{y})$
Differentiating with respect to x,
$\frac{\text{d}}{\text{dx}}\big[\text{y}\log\sin\text{x}\big]=\frac{\text{d}}{\text{dx}}\big[\text{x}\log\cos\text{y}\big]$
$\Rightarrow\text{y}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}} \\ =\text{x}\frac{\text{dy}}{\text{dx}}(\log\cos\text{y})+\log\cos\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\text{y}\Big(\frac{1}{\sin\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}} \\ =\frac{\text{x}}{\cos\text{y}}\frac{\text{x}}{\text{dx}}(\cos\text{y})+\log\cos\text{y}(1)$
$\Rightarrow\frac{\text{y}}{\sin\text{x}}(\cos\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}} \\ =\frac{\text{x}}{\cos\text{y}}(-\sin\text{y})\frac{\text{dy}}{\text{dx}}+\log\cos\text{y}$
$\Rightarrow \text{y}\cot\text{x}+\log\sin\text{x}\frac{\text{dy}}{\text{dx}} \\ =-\text{x}\tan\text{y}\frac{\text{dy}}{\text{dx}}+\log\cos\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(\log\sin\text{x}+\text{x}\tan\text{y}) \\ =\log\cos\text{y}-\text{y}\cot\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\log\cos\text{y}-\text{y}\cot\text{x}}{\log\sin\text{x}+\text{x}\tan\text{y}}$
View full question & answer→Question 4055 Marks
If $\text{x}=3\sin\text{t}-\sin3\text{t},$ $\text{y}=3\cos-\cos3\text{t}$ find $\frac{\text{dy}}{\text{dx}}$ at $\text{t}=\frac{\pi}{3}.$
Answer$\text{x}=3\sin\text{t}-\sin3\text{t},$ $\text{y}=3\cos\text{t}-\cos3\text{t},$
$\therefore\ \frac{\text{dx}}{\text{dt}}=3.\frac{\text{d}}{\text{dt}}\sin\text{t}-\frac{\text{d}}{\text{dx}}\sin3\text{t}$
$=3\cos\text{t}-\cos3\text{t}.\frac{\text{d}}{\text{dt}}3\text{t}=\cos\text{t}-3\cos3\text{t}\ \ \dots(\text{i})$
and $\frac{\text{dy}}{\text{dt}}=3.\frac{\text{d}}{\text{dt}}\cos\text{t}-\frac{\text{d}}{\text{dt}}\cos3\text{t}$
$=3\sin\text{t}+\sin3\text{t}.\frac{\text{d}}{\text{dt}}3\text{t}$
$\frac{\text{dy}}{\text{dt}}=3\sin3\text{t}-3\text{t}\sin\text{t}\ \ \dots(\text{ii})$
$\therefore\ \frac{\text{dy}}{\text{dt}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{3(\sin3\text{t}-\sin\text{t})}{3(\cos\text{t}-\cos3\text{t})}$
Now, $\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{t}=\frac{\pi}{3}}=\frac{\sin\frac{3\pi}{3}-\sin\frac{\pi}{3}}{\Big(\cos\frac{\pi}{3}-\cos3\frac{\pi}{3}\Big)}$ $=\frac{0-\frac{\sqrt{3}}{2}}{\frac{1}{2}-(-1)}$
$=\frac{-\frac{\sqrt{3}}{2}}{\frac{3}{2}}=\frac{-\sqrt{3}}{3}=\frac{-1}{\sqrt{3}}$
View full question & answer→Question 4065 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\sqrt{1+\text{a}^2\text{x}^2-1}}{\text{ax}}\Big),\text{x}\neq0$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\text{a}^2\text{x}^2}-1}{\text{ax}}\Big)$
Put $\text{ax}=\tan\theta$
$\text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\text{a}^2\text{x}^2}-1}{\text{ax}}\Big)$
$=\tan^{-1}\Big(\frac{\sec\theta-1}{\tan\theta}\Big)$
$=\tan^{-1}\Big(\frac{1-\cos\theta}{\sin\theta}\Big)$
$=\tan^{-1}\bigg(\frac{\frac{2\sin^2\theta}{2}}{\frac{2\sin\theta}{2}\frac{\cos\theta}{2}}\bigg)$
$\text{y}=\tan^{-1}\Big(\frac{\tan\theta}{2}\Big)$
$=\frac{\theta}{2}$
$\text{y}=\frac{1}{2}\tan^{-1}(\text{ax})$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\times\Big(\frac{1}{1+(\text{ax})^2}\Big)\frac{\text{d}}{\text{dx}}(\text{ax})$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2(1+\text{a}^2\text{x}^2)}(\text{a})$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{a}}{2(1+\text{a}^2\text{x}^2)}$
View full question & answer→Question 4075 Marks
Differentiate the following functions with respect to x:
$\log\sqrt{\frac{\text{x}-1}{\text{x}+1}}$
AnswerLet $\text{y}=\log\sqrt{\frac{\text{x}-1}{\text{x}+1}}$
$\Rightarrow\text{y}=\log\Big(\frac{\text{x}-1}{\text{x}+1}\Big)^\frac{1}{2}$
$\Rightarrow\text{y}=\frac{1}{2}\log\Big(\frac{\text{x}-1}{\text{x}+1}\Big)$
$\Rightarrow\text{y}=\frac{1}{2}\big[\log(\text{x}-1)-\log(\text{x}+1)\big]$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\Big[\frac{\text{d}}{\text{dx}}\big\{\log(\text{x}-1)\big\}-\frac{\text{d}}{\text{dx}}\big\{\log(\text{x}+1)\big\}\Big]$
$=\frac{1}{2}\Big(\frac{1}{\text{x}-1}-\frac{1}{\text{x}+1}\Big)$
$=\frac{1}{2}\Big(\frac{2}{\text{x}^2-1}\Big)$
$=\frac{2}{\text{x}^2-1}$
So,
$\frac{\text{dy}}{\text{dx}}=\frac{2}{\text{x}^2-1}$
View full question & answer→Question 4085 Marks
Differentiate the following functions from first principles:
$x^2e^x$.
AnswerLet $f(x) = x^2e^x$
$\Rightarrow f(x + h) = (x + h)^2 e^{(x+h)}$
$\therefore \frac{\text{d}}{\text{dx}}\{\text{f(x)}\}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{x}+\text{h})^2\text{e}^{(\text{x}+\text{h})}-\text{x}^2\text{e}^{\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\text{x}^2\text{e}^{(\text{x}+\text{h})}-\text{x}^2\text{e}^\text{x}}{\text{h}}+\frac{2\text{xhe}^{(\text{x}+\text{h})}}{\text{h}}+\frac{\text{h}^2\text{e}^{(\text{x}+\text{h})}}{\text{h}}\Big)$
$=\lim\limits_{\text{x}\rightarrow0}\bigg(\frac{\text{x}^2\text{e}^\text{x}\big(\text{e}^{(\text{x}+\text{h}-\text{x})}-1\big)}{\text{x}}+2\text{xe}^{(\text{x}+\text{h})}+\text{he}^{(\text{x}+\text{h})}\bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[\text{x}^2\text{e}^{\text{x}}\frac{\big(\text{e}^\text{h}-1\big)}{\text{h}}+2\text{xe}^{(\text{x}+\text{h})}+\text{h}^{\text{e}}(\text{x}+\text{h})\bigg]$
$=\text{x}^2\text{e}^\text{x}+2\text{xe}^\text{x}+0\times\text{e}^\text{x}\ \Big[\text{Since,}\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}=1\Big]$
So,
$\frac{\text{d}}{\text{dx}}(\text{x}^2\text{e}^\text{x})=\text{e}^\text{x}(\text{x}^2+2\text{x})$
View full question & answer→Question 4095 Marks
If $\log\sqrt{\text{x}^2+\text{y}^2}=\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big),$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\text{x}+\text{y}}{\text{x}-\text{y}}$
AnswerHere,
$\log\sqrt{\text{x}^2+\text{y}^2}=\tan^{-1}\Big(\frac{\text{x}}{\text{y}}\Big)$
$\Rightarrow\log(\text{x}^2+\text{y}^2)^{\frac{1}{2}}=\tan^{-1}\Big(\frac{\text{x}}{\text{y}}\Big)$
$\Rightarrow\frac{1}{2}\log(\text{x}^2+\text{y}^2)=\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)$
Differentiating with respect to x,
$\Rightarrow\frac{1}{2}\frac{\text{d}}{\text{dx}}\log(\text{x}^2+\text{y}^2)=\frac{\text{d}}{\text{dx}}\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)$
$\Rightarrow\frac{1}{2}\times\Big(\frac{1}{\text{x}^2+\text{y}^2}\Big)\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{y}^2)=\frac{1}{1+\Big(\frac{\text{y}}{\text{x}}\Big)^2}\frac{\text{d}}{\text{dx}}\Big(\frac{\text{y}}{\text{x}}\Big)$
$\Rightarrow\frac{1}{2}\Big(\frac{1}{\text{x}^2+\text{y}^2}\Big)\Big[2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big]=\frac{\text{x}^2}{(\text{x}^2+\text{y}^2)}\bigg[\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\frac{\text{d}}{\text{dx}}(\text{x})}{\text{x}^2}\bigg]$
$ \Rightarrow\frac{1}{2}\Big(\frac{1}{\text{x}^2+\text{y}^2}\Big)\times2\Big(\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{x}^2}{(\text{x}^2+\text{y}^2)}\bigg[\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}(1)}{\text{x}^2}\bigg]$
$\Rightarrow\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}-\text{x}\frac{\text{dy}}{\text{dx}}=-\text{y}-\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(\text{y}-\text{x})=-(\text{y}+\text{x})$
View full question & answer→Question 4105 Marks
Find $\text{y}=\text{Ae}^{-\text{kt}}\cos\text({pt}+\text{c})$prove that $\frac{\text{d}^2\text{y}}{\text{dt}^2}+2\text{k}\frac{\text{dy}}{\text{dt}}+\text{n}^2\text{y}=0,$Where $\text{n}^2=\text{p}^2+\text{k}^2.$
AnswerWe have,
$\text{y}=\text{Ae}^{-\text{kt}}\cos\text({pt}+\text{c})...(1)$
Differentiating y with respect to t, we get
$\frac{\text{dy}}{\text{dt}}=-\text{KAe}^{-\text{kt}}\cos(\text{pt}+\text{c})-\text{PAe}^{-\text{kt}}\sin(\text{pt}+\text{c})$
$=-\text{ky}-\text{PAe}^{-\text{kt}}\sin(\text{pt}+\text{c})\ [\text{from}(1)]$
$\Rightarrow\text{pAe}^{-\text{kt}}\sin(\text{pt}+\text{c})=-\text{ky}-\frac{\text{dy}}{\text{dt}}...(2)$
Differentiating $\frac{\text{dy}}{\text{dt}}$ with respect to t, we get
$\frac{\text{d}^2\text{y}}{\text{dt}}=-\text{k}\frac{\text{dy}}{\text{dt}}+\text{pkAe}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{p}^2\text{Ae}^{-\text{kt}}\cos(\text{pt}+\text{c})$
$=-\text{k}\frac{\text{dy}}{\text{dt}}+\text{k}\Big(-\text{ky}-\frac{\text{dy}}{\text{dt}}\Big)-\text{p}^2\text{y}\ [\text{from}(1)\ \text{and}\ (2)]$
$=-\text{k}\frac{\text{dy}}{\text{dt}}-\text{k}^2\text{y}-\text{k}\frac{\text{dy}}{\text{dt}}-\text{p}^2\text{y}$
$=-2\text{k}\frac{\text{dy}}{\text{dt}}-(\text{k}^2+\text{p}^2)\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}+2\text{k}\frac{\text{dy}}{\text{dt}}+(\text{k}^2+\text{p}^2)\text{y}=0$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}+2\text{k}\frac{\text{dy}}{\text{dt}} \text{n}^2\text{y}=0,$ where $\text{n}^2=\text{p}^2+\text{k}^2.$
Hence,
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}+2\text{k}\frac{\text{dy}}{\text{dt}} \text{n}^2\text{y}=0,$ where $\text{n}^2=\text{p}^2+\text{k}^2.$
View full question & answer→Question 4115 Marks
Differentiate the following functions with respect to x:
$(\cos\text{x})^\text{x}+(\sin\text{x})^\frac{1}{\text{x}}$
AnswerLet $\text{y}=(\cos\text{x})^\text{x}+(\sin\text{x})^\frac{1}{\text{x}}$
$\Rightarrow\text{y}=\text{e}^{\log(\cos\text{x})^\text{x}}+\text{e}^{\log(\sin\text{x})^\frac{1}{\text{x}}}$
$\Rightarrow\text{y}=\text{e}^{\text{x}\log(\cos\text{x})}+\text{e}^{\frac{1}{\text{x}}\log\sin\text{x}}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\cos\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\frac{1}{\text{x}}\log\sin\text{x}}\big)$
$=\text{e}^{\log\cos\text{x}}\times\frac{\text{d}}{\text{dx}}(\text{x}\log\cos\text{x})+\text{e}^{\frac{1}{\text{x}}\log\sin}\frac{\text{d}}{\text{dx}}\big(\frac{1}{\text{x}}\log\sin\text{x}\big)$
$=\text{e}^{\log(\cos\text{x})^\text{x}}\times\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\cos\text{x})+\log\cos\text{x}\times\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +\text{e}^{\log(\sin\text{x})^\frac{1}{\text{x}}}\times\Big[\frac{1}{\text{x}}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{d}}{\text{dx}}\big(\frac{1}{\text{x}}\big)\Big]$
$=(\cos\text{x})^\text{x}\Big[\text{x}\big(\frac{1}{\cos\text{x}}\big)\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}+\log\cos\text{x}(1)\Big] \\ +(\sin)^\frac{1}{\text{x}}\Big[\frac{1}{\text{x}}\times\frac{1}{\sin\text{x}}\times\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\Big(-\frac{1}{\text{x}^2}\Big)\Big]$
$=(\cos\text{x})^\text{x}\Big[\text{x}\Big(\frac{1}{\cos\text{x}}\Big)(-\sin\text{x})+\log\cos\text{x}\Big] \\ +(\sin\text{x})^\frac{1}{\text{x}}\Big[\frac{1}{\text{x}}\times\frac{1}{\sin\text{x}}(\cos\text{x})-\frac{1}{\text{x}^2}\log\sin\text{x}\Big]$
$=(\cos\text{x})^\text{x}\big[\log\cos\text{x}-\text{x}\tan\text{x}\big](\sin\text{x})^\frac{1}{\text{x}} \\ \Big[\frac{\cot\text{x}}{\text{x}}-\frac{1}{\text{x}^2}\log\sin\text{x}\Big]$
View full question & answer→Question 4125 Marks
Differentiate the following functions with respect to x:
$\text{e}^\text{x}\log\sin2\text{x}$
AnswerLet $\text{y}=\text{e}^\text{x}\log\sin2\text{x}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\text{e}^\text{x}\log\sin2\text{x}\big]$
$=\text{e}^\text{x}\frac{\text{d}}{\text{dx}}\log\sin2\text{x}+\log\sin2\text{x}\frac{\text{d}}{\text{dx}}\big(\text{e}^\text{x}\big)$
[Using product rule and chain rule]
$=\text{e}^\text{x}\frac{1}{\sin2\text{x}}\frac{\text{d}}{\text{dx}}(\sin2\text{x})+\log\sin2\text{x}\big(\text{e}^\text{x}\big)$
$=\frac{\text{e}^\text{x}}{\sin2\text{x}}\cos2\text{x}\frac{\text{d}}{\text{dx}}(2\text{x})+\text{e}^\text{x}\log\sin2\text{x}$
$=\frac{2\cos2\text{xe}^\text{x}}{\sin2\text{x}}+\text{e}^\text{x}\log\sin2\text{x}$
$=\text{e}^\text{x}(2\cot2\text{x}+\log\sin2\text{x})$
So,
$\frac{\text{d}}{\text{dx}}\big(\text{e}^\text{x}\log\sin2\text{x}\big)=\text{e}^\text{x}(2\cot2\text{x}+\log\sin2\text{x})$
View full question & answer→Question 4135 Marks
Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}|\text{x}-\text{a|}\sin(\frac{1}{\text{x}-\text{a}}), &\text{for} \text{ x} \neq\text{a}\\0,&\text{for} \text{ x} = \text{a}\end{cases}\text{ at x}=0$
Answer$\text{f}\text{(x)}=\begin{cases}|\text{x}-\text{a|}\sin\Big(\frac{1}{\text{x}-\text{a}}\Big),&\text{for }\text{x }\neq \text{a}\\0, &\text{for x } = \text{a}\end{cases}$ $(\text{LHL at x}=\text{a})\lim\limits_{\text{x} \rightarrow \text{a}^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow \text{0}}\text{f}\text{(a}-\text{h)}$ $=\lim\limits_{\text{h} \rightarrow0}\text{|a}-\text{h}-\text{a|}\sin\Big(\frac{1}{\text{a}-\text{h}-\text{a}}\Big)=\lim\limits_{\text{h} \rightarrow0}\text{h}\sin\Big(\frac{1}{-\text{h}}\Big)$= 0 × (an oscillating number between -1 and 1)
= 0 $(\text{RHL at x}=\text{a})\lim\limits_{\text{x} \rightarrow \text{a}^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow \text{0}}\text{f}\text{(a}+\text{h)}$ $=\lim\limits_{\text{h} \rightarrow0}\text{|a}+\text{h}-\text{a|}\sin\Big(\frac{1}{\text{a}+\text{h}-\text{a}}\Big)=\lim\limits_{\text{h} \rightarrow0}\text{h}\sin\Big(\frac{1}{\text{h}}\Big)$= 0 × (an oscillating number between -1 and 1)
Thus, we obtian $\lim\limits_{\text{x} \rightarrow\text{a}^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow\text{a}^+}\text{f}\text{(x)}=\text{f}\text{(a)}$ $\therefore$ f(x) is continuous at x = a.
View full question & answer→Question 4145 Marks
If $\text{y}\sqrt{\text{x}^2+1}=\log\Big(\sqrt{\text{x}^2+1}-\text{x}\Big),$ prove that $\big(\text{x}^2+1\big)\frac{\text{dx}}{\text{dx}}+\text{xy}+1=0$
AnswerWe have, $\text{y}\sqrt{\text{x}^2+1}=\log\Big(\sqrt{\text{x}^2+1}-\text{x}\Big)$Differentiating with respect to x, we get,
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{y}\sqrt{\text{x}^2+1}\Big)=\frac{\text{d}}{\text{dx}}\log\Big(\sqrt{\text{x}^2+1}-\text{x}\Big)$
[Using Product rule and chain rule]
$\Rightarrow\text{y}\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}^2+1}\big)+\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}} \\ =\frac{1}{\big(\sqrt{\text{x}^2+1}-\text{x}\big)}\times\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}^2+1}-\text{x}\big)$
$\Rightarrow\frac{\text{y}}{2\sqrt{\text{x}^2+1}}\times\frac{\text{d}}{\text{dx}}(\text{x}^2+1)+\sqrt{\text{x}^2+1}\frac{\text{d}}{\text{dx}} \\ =\frac{1}{\big(\sqrt{\text{x}^2+1}-\text{x}\big)}\times\Big[\frac{1}{2\sqrt{\text{x}^2+1}}\frac{\text{d}}{\text{dx}}(\text{x}^2+1)-1\Big]$
$\Rightarrow\ \frac{2\text{xy}}{2\sqrt{\text{x}^2+1}}+\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}} \\ =\frac{1}{\big(\sqrt{\text{x}^2+1}-\text{x}\big)}\Big[\frac{2\text{x}}{2\sqrt{\text{x}^2+1}}-1\Big]$
$\Rightarrow\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}} \\ =\Big[\frac{1}{\sqrt{\text{x}^2+1}-\text{x}}\Big]\Big[\frac{\text{x}-\sqrt{\text{x}^2+1}}{\sqrt{\text{x}^2+1}}\Big]-\frac{\text{xy}}{\sqrt{\text{x}^2+1}}$
$\Rightarrow\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}}=\frac{-(1+\text{xy})}{\sqrt{\text{x}^2+1}}$
$\Rightarrow\big(\text{x}^2+1\big)\frac{\text{dy}}{\text{dx}}=-(1+\text{xy})$
$\Rightarrow(\text{x}^2+1)\frac{\text{dy}}{\text{dx}}+1+\text{xy}=0$
View full question & answer→Question 4155 Marks
If $y^x = e^{y-x}, $Prove that $\frac{\text{dy}}{\text{dx}}=\frac{(1+\log\text{y})^2}{\log\text{y}}$
AnswerHere,
$y^x = e^{y-x}$
Taking log on both the sides,
$\log\text{y}=\log\text{e}^{(\text{y}-\text{x})}$
$\big[\text{Since},\log\text{a}^{\text{b}}=\text{b}\log\text{a and}\log_\text{e}\text{e}=1\big]$
$\text{x}\log\text{y}=(\text{y}-\text{x})\log\text{e}$
$\text{x}\log\text{y}=\text{y}-\text{x}\ .....(\text{i})$
Differentiating it with respect to x using product rule,
$\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})=\frac{\text{d}}{\text{dx}}(\text{y}-\text{x})$
$\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big]=\frac{\text{dy}}{\text{dx}}-1$
$\text{x}\Big(\frac{\text{x}}{\text{y}}\Big)\frac{\text{dy}}{\text{dx}}+\log\text{y}(1)=\frac{\text{dy}}{\text{dx}}-1$
$\frac{\text{dx}}{\text{dx}}\Big(\frac{\text{x}}{\text{y}}-1\Big)=-1-\log\text{y}$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{\text{y}}{(1+\log\text{y})\text{y}}\Big)=-(1+\log\text{y})$
$\Big[\text{Since, from equation (i), x}=\frac{\text{y}}{(1+\log\text{y})}\Big]$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{1-1-\log\text{y}}{(1+\log\text{y})}\Big]=-(1+\log\text{y})$
$\frac{\text{dy}}{\text{dx}}=-\frac{(1+\log\text{y})^2}{-\log\text{y}}$
$\frac{\text{dy}}{\text{dx}}=\frac{(1+\log\text{y})^2}{\log\text{y}}$
View full question & answer→Question 4165 Marks
Show that the function $\text{f(x)}\begin{cases}\text{x}^\text{m}\sin(\frac{1}{\text{x}}), &\text{x}\neq0 \\0 ,& \text{x}=0\end{cases}$
Neirher continuous but not diffierentiable, if $\text{m}\leq0$
Answer$\text{LHL }=\lim_\limits{\text{x}\rightarrow0^{-}}\text{f(x)}$ $=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$ $=\lim_\limits{\text{h}\rightarrow0}(-\text{h})^\text{m}\sin\Big(-\frac{1}{\text{h}}\Big)$ = Not defined as $\text{m}\leq0$ $\text{RHL }=\lim_\limits{\text{x}\rightarrow0^{+}}\text{f(x)}$ $=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$ $=\lim_\limits{\text{h}\rightarrow0}(+\text{h})^\text{m}\sin\Big(\frac{1}{0+\text{h}}\Big)$ =Not defined, as $\text{m}\leq0$ Since RHL and LHL are not difined, so f(x) is not continuous Let x = 0 for $\text{m}\leq0.$ (LHL at x = 0) $=\lim_\limits{\text{x}\rightarrow0^{-1}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$ $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0-\text{h})-(0)}{0-\text{h}-0}$ $=\lim_\limits{\text{h}\rightarrow0}\frac{(-\text{h})^\text{m}\sin\Big(-\frac{1}{\text{h}}\Big)}{-\text{h}}$ $=\lim_\limits{\text{h}\rightarrow0}(-\text{h})^\text{m-1}\sin\Big(-\frac{1}{\text{h}}\Big)$ = Not definded, as $\text{m}\leq0$ (RHL at x = 0) $=\lim_\limits{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$ $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0+\text{h})-\text{f}(0)}{0+\text{h}-0}$ $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^\text{m}\sin\Big(\frac{1}{\text{h}}\Big)}{\text{h}}$ $=\lim_\limits{\text{h}\rightarrow0}(\text{h})^{\text{m}^{-1}}\sin\Big(\frac{1}{\text{h}}\Big)$ = Not defined $\text{m}\leq0$ Thus,f(x) is neither continuous not differentiable at x = 0 for $\text{m}\leq0.$
View full question & answer→Question 4175 Marks
Show that the function f defined as follows,
$\text{f(x)}=\begin{cases}3\text{x}-2, & 0<\text{x}\leq1\\2\text{x}^2-\text{x,} & 1<\text{x}\leq2\\5\text{x}-4,&\text{x}>2\end{cases}$
is countinuous at x = 2, but not differentiable there at x = 2.
AnswerGiven:
$\text{f(x)}=\begin{cases}3\text{x}-2, & 0<\text{x}\leq1\\2\text{x}^2-\text{x,} & 1<\text{x}\leq2\\5\text{x}-4,&\text{x}>2\end{cases}$
First, we will show that f(x) is continuos at x = 2.
We have,
(LHL at x = 2)
$=\lim_\limits{\text{x}\rightarrow2^{-}}\text{f(x)}$
$=\lim_\limits{\text{x}\rightarrow0}\text{f}(2-\text{h)}$
$=\lim_\limits{\text{x}\rightarrow0}2(2-\text{h)}^2-(2-\text{h})$
$=\lim_\limits{\text{x}\rightarrow0}(8+2\text{h}^2-8\text{h}-2+\text{h})$
$=6$
(RHL at x = 2)
$=\lim_\limits{\text{x}\rightarrow2^{+}}\text{f(x)}$
$=\lim_\limits{\text{x}\rightarrow0}\text{f}(2+\text{h)}$
$=\lim_\limits{\text{x}\rightarrow0}5(2+\text{h)}-4$
$=\lim_\limits{\text{x}\rightarrow0}(10+5\text{h}-4)$
$=6$
and $\text{f}(2)=2\times4-2=6$
Thus, $=\lim_\limits{\text{x}\rightarrow2^{-}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow2^{+}}\text{f(x)}=\text{f}(2)$
Hence the function is continuous at x = 2.
Now, we will check whether the given function is differerentiable at x = 2.
We have,
(LHL at x = 2)
$\lim_\limits{\text{x}\rightarrow2^{-}}\frac{\text{f(x)}-\text{f}(2)}{\text{x}-2}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\text{f}(2-\text{h})-\text{f}(2)}{-\text{h}}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{2\text{h}^2-7\text{h}+6-6}{-\text{h}}$
$=\lim_\limits{\text{x}\rightarrow0}-2\text{h}+7$
$=7$
(RHL at x = 2)
$\lim_\limits{\text{x}\rightarrow2^{+}}\frac{\text{f(x)}-\text{f}(2)}{\text{x}-2}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\text{f}(2+\text{h})-\text{f}(2)}{\text{h}}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{10+5\text{h}-4-6}{\text{h}}$
$=5$
Thus, LHL at x = 2 $\neq$ RHL at x = 2.
Hence, function is no differentiable at x = 2.
View full question & answer→Question 4185 Marks
Differentiate the following functions with respect to x:
$\log(3\text{x}+2)-\text{x}^2\log(2\text{x}-1)$
AnswerLet $\text{y}=\log(3\text{x}+2)-\text{x}^2\log(2\text{x}-1)$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\log(3\text{x}+2)-\text{x}^2\log(2\text{x}-1)\big]$
$\frac{\text{d}}{\text{dx}}\log(3\text{x}+2)-\frac{\text{d}}{\text{dx}}\big(\text{x}^2\log(2\text{x}-1)\big)$
$=\frac{1}{3\text{x}+2}\frac{\text{d}}{\text{dx}}(3\text{x}+2)-\Big[\text{x}^2\frac{\text{d}}{\text{dx}}\log(2\text{x}-1)+\log(2\text{x}-1)\frac{\text{d}}{\text{dx}}\big(\text{x}^2\big)\Big]$
[Using product rule and chain rule]
$=\frac{3}{3\text{x}+2}\Big[\text{x}^2\times\frac{1}{2\text{x}-1}\frac{\text{d}}{\text{dx}}(2\text{x}-1)+\log(2\text{x}-1)\times2\text{x}\Big]$
$=\frac{3}{3\text{x}+2}-\frac{2\text{x}^2}{2\text{x}-1}-2\text{x}\log(2\text{x}-1)$
So,
$\frac{\text{d}}{\text{dx}}\big(\log(3\text{x}+2)-\text{x}^2\log(2\text{x}-1)\big) \\ =\frac{3}{3\text{x}+2}-\frac{2\text{x}^2}{2\text{x}-1}-2\text{x}\log(2\text{x}-1)$
View full question & answer→Question 4195 Marks
If $x^x + y^x = 1$, prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}(\text{y}+\text{x}\log\text{y})}{\text{x}(\text{y}\log\text{x}+\text{x})}$
AnswerHere,
$x^x + y^x = 1$
Taking on bith sides,
$\log(\text{x}^\text{y}\times\text{y}^\text{x})=\log(1)$
$\text{y}=\log\text{x}+\text{x}\log\text{y}=\log1$
$\big[\text{Since}, \log(\text{AB})=\log\text{A}+\log\text{B},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using product rule,
$\frac{\text{d}}{\text{dx}}(\text{y}\log\text{x})+\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})=\frac{\text{d}}{\text{dx}}(\log1)$
$\Big[\text{y}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{dy}}{\text{dx}}\Big]+\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}\text{(x)}\Big]=0$
$\Big[\text{y}\Big(\frac{1}{\text{x}}\Big)\log\text{x}\frac{\text{dy}}{\text{dx}}\Big]+\Big[\text{x}\Big(\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}\Big)+\log\text{y}(1)\Big]=0$
$\frac{\text{y}}{\text{x}}+\log\text{x}\log\frac{\text{dy}}{\text{dx}}+\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}=0$
$\frac{\text{dy}}{\text{dx}}\Big(\log\text{x}+\frac{\text{x}}{\text{y}}\Big)=-\Big[\log\text{y}+\frac{\text{y}}{\text{x}}\Big]$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{y}\log\text{x}+\text{x}}{\text{y}}\Big]=-\Big[\frac{\text{x}\log\text{y}+\text{y}}{\text{x}}\Big]$
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}\Big[\frac{\text{x}\log\text{y}+\text{y}}{\text{y}\log\text{x}+\text{x}}\Big]$
View full question & answer→Question 4205 Marks
If $\text{x}=\text{a}\sin2\text{t}(1+\cos 2\text{t})$ and $\text{y}=\text{b}\cos\text{t}(1-\cos2\text{t}),$ show that at $\text{t}=\frac{\pi}{4},\frac{\text{dy}}{\text{dx}}=\frac{\text{b}}{\text{a}}\text{ t}=\frac{\pi}{4},\frac{\text{dy}}{\text{dx}}=\frac{\text{b}}{\text{a}}$
AnswerConsider the given functions,
$\text{x}=\text{a}\sin(2\text{t})(1+\cos2\text{t})\text{ and y}=\text{b}\cos2\text{t}(1-\cos2\text{t})$
Rewriting the above function, we have,
$\text{x}=\text{a}\sin2\text{t}+\frac{\text{a}}{2}\sin4\text{t}$
Differentiating the above function w.r.t. 't', we have,
$\frac{\text{dx}}{\text{dx}}=2\text{a}\cos2\text{t}+2\text{a}\cos4\text{t}\ .....(\text{i})$
$\text{y}=\text{b}\cos2\text{t}(1-\cos2\text{t})$
$\text{y}=\text{b}\cos2\text{t}-\text{b}\cos^22\text{t}$
$\frac{\text{dy}}{\text{dt}}=-2\text{b}\sin2\text{t}+2\text{b}\cos2\text{t}\sin2\text{t} \\ =-2\text{b}\sin2\text{t}+\text{b}\sin4\text{t}\ .....(\text{ii})$
From (1) and (2),
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-2\text{b}\sin2\text{t}+\text{b}\sin4\text{t}}{2\text{a}\cos2\text{t}+2\text{a}\cos4\text{t}}$
$\therefore\frac{\text{dy}}{\text{dx}}\Big|_{\frac{\pi}{4}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}\Bigg|_{\text{t}=\frac{\pi}{4}}=\frac{-2\text{b}}{-2\text{a}}=\frac{\text{b}}{\text{a}}$
View full question & answer→Question 4215 Marks
If $\text{y}=\text{x}\sin(\text{a}+\text{y}),$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})-\text{y}\cos(\text{a}+\text{y})}$
AnswerWe have, $\text{y}=\text{x}\sin(\text{a}+\text{y})$
Differentiate with respect to y,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\text{x}\sin(\text{a}+\text{y})\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\{\sin(\text{a}+\text{y})\}+\sin(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})$
[Using product rule and chain rule]
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\cos(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}+\sin(\text{a}+\text{y})(1)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\{1-\text{x}\cos(\text{a}+\text{y})\}=\sin(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin(\text{a}+\text{y})}{1-\text{x}\cos(\text{a}+\text{y})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin(\text{a}+\text{y})}{1-\frac{\text{y}}{\sin(\text{a}+\text{y})}\cos(\text{a}+\text{y})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y}){-\text{y}\cos(\text{a}+\text{y})}}$
Hence, proved.
View full question & answer→Question 4225 Marks
Find $\frac{\text{dy}}{\text{dx}},$ if $\text{y}=\text{x}^{\tan\text{x}}+\sqrt{\frac{\text{x}^2+1}{2}}.$
AnswerWe have, $\text{y}=\text{x}^{\tan\text{x}}+\sqrt{\frac{\text{x}^2+1}{2}}$
Let $\text{u}=\text{x}^{\tan\text{x}}$ and $\text{v}=\sqrt{\frac{\text{x}^2+1}{2}}$
$\therefore\ \log\text{u}\log\text{x}^{\tan\text{x}}=\tan\text{x}\log\text{x}$
Differentiating w.r.t. x, we get
$\frac{1}{\text{u}}\cdot\frac{\text{du}}{\text{dx}}=\tan\text{x}\cdot\frac{1}{\text{x}}+\log\text{x}\cdot\sec^2\text{x}$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{u}\Big[\frac{\tan\text{x}}{\text{x}}+\log\text{x}\cdot\sec^2\text{x}\Big]$
$=\text{x}^{\tan\text{x}}\Big[\frac{\tan\text{x}}{\text{x}}+\log\text{x}\cdot\sec^2\text{x}\Big]$
Now, $\text{v}=\sqrt{\frac{\text{x}^2+1}{2}}$
Differentiating w.r.t. x, we get
$\frac{\text{dv}}{\text{dx}}=\frac{1}{2}\Big(\frac{\text{x}^2+1}{2}\Big)^{\frac{-1}{2}}\frac{2\text{x}}{2}=\frac{\text{x}}{2}\sqrt{\frac{2}{\text{x}^2+1}}$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=\frac{\text{x}}{\sqrt{2(\text{x}^2+1)}}$
Now, $\text{y}=\text{u}+\text{v}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}$ $=\text{x}^{\tan\text{x}}\Big[\frac{\tan\text{x}}{\text{x}}+\log\text{x}\cdot\sec^2\text{x}\Big]+\frac{\text{x}}{\sqrt{2(\text{x}^2+1)}}$
View full question & answer→Question 4235 Marks
Let C be a curve defined parametrically as $\text{x}=\text{a}\cos^3\theta,\text{y}=\text{a}\sin^3\theta,0\leq\theta\leq\frac{\pi}{2}.$ Determine a point P on C, where the tangent to C is parallel to the chord joining the points (a, 0) and (0, a).
AnswerAs, $\text{x}=\text{a}\cos^3\theta$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=-3\text{a}\cos^2\theta\sin\theta$
And, $\text{y}=\text{a}\sin^3\theta$
$\Rightarrow\frac{\text{dy}}{\text{d}\theta}=3\text{a}\sin^2\theta\cos\theta$
So, $\frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{d}\theta}\Big)}{\Big(\frac{\text{dx}}{\text{d}\theta}\Big)}=\frac{3\text{a}\sin^2\theta\cos\theta}{-3\text{a}\cos^2\theta\sin\theta}=-\tan\theta$
For the tangent to be parallel to the chord joining the points (a, 0) and (0, a).
$\frac{\text{dy}}{\text{dx}}=\frac{0-\text{a}}{\text{a}-0}$
$\Rightarrow-\tan\theta=-1$
$\Rightarrow\tan\theta=1$
$\Rightarrow\theta=\frac{\pi}{4}$
Now,
$\text{x}=\text{a}\cos^3\frac{\pi}{4}=\text{a}\Big(\frac{1}{\sqrt{2}}\Big)^3=\frac{\text{a}}{2\sqrt2}$ and
$\text{y}=\text{a}\sin^3\frac{\pi}{4}=\text{a}\Big(\frac{1}{\sqrt{2}}\Big)^3=\frac{\text{a}}{2\sqrt2}$
So, the point P on the curve C is $\Big(\frac{\text{a}}{2\sqrt2},\frac{\text{a}}{2\sqrt2}\Big).$
View full question & answer→Question 4245 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$y = e^x + 10^x + x^x$
AnswerHere, $\text{y}=\text{e}^\text{x}+10^\text{x}+\text{x}^\text{x}$
$\text{e}^{\text{x}}+10^{\text{x}}+\text{e}^{\log\text{x}^\text{x}}$
$\big[\text{Since, e}^{{\log}_\text{a}^\text{b}}=\text{a},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
$\text{y}=\text{e}^{\text{x}}+10^{\text{x}}+\text{e}^{\log\text{x}^\text{x}}$
Differentiating it with respect to x using product rule, chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})+\frac{\text{d}}{\text{dx}}(10^{\text{x}})+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{x}}\big)$
$=\text{e}^\text{x}+10^\text{x}\log10+\text{e}^{\text{x}\log\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})$
$=\text{e}^\text{x}+10^\text{x}\log10+\text{e}^{\text{x}\log\text{x}}\Big[\text{x}\times\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$=\text{e}^\text{x}+10^\text{x}\log10+\text{e}^{\text{x}\log\text{x}}\Big[\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)\Big]$
$=\text{e}^\text{x}+10^\text{x}\log10+\text{x}^{\text{x}}[1+\log\text{x}]$
$=\text{e}^\text{x}+10^\text{x}\log10+\text{x}^{\text{x}}[\log\text{e}+\log\text{x}] \big[\text{Since}, \log_\text{e}\text{e}=1\big]$
$\frac{\text{dy}}{\text{dx}}=\text{e}^\text{x}+10^\text{x}\log10+\text{x}^\text{x}(\log\text{ex})\ \big[\text{Since}\log\text{A}+\log\text{B}=\log\text{AB}]$
View full question & answer→Question 4255 Marks
If for function $\phi(\text{x})=\lambda\text{x}^2+7\text{x}-4, \phi(5)=97,$ find $\lambda.$
AnswerGiven: $\phi(\text{x})=\lambda\text{x}^2+7\text{x}-4$
Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of $\phi$ at x is given by:
$\phi'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\phi(\text{x}+\text{h})-\phi(\text{x})}{\text{h}}$
$\Rightarrow\phi'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\lambda(\text{x}+\text{h})^2+7(\text{x}+\text{h})-4-\lambda\text{x}^2-7\text{x}+4}{\text{h}}$
$\Rightarrow\phi'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\lambda(\text{x}+\text{h})^2+7(\text{x}+\text{h})-4-\lambda\text{x}^2-7\text{x}+4}{\text{h}}$
$\Rightarrow\phi'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\lambda\text{h}^2+2\lambda\text{xh}+7\text{h}}{\text{h}}$
$\Rightarrow\phi'(\text{x})=\lim_\limits{\text{h}\rightarrow\infty}\frac{\text{h}(\lambda\text{h}+2\lambda\text{x}+7)}{\text{h}}$
$\Rightarrow\phi'(\text{x})=2\lambda\text{x}+7$
It is given $\phi'(5)=97$
Thus, $\phi'(5)=10\lambda+7=97$
$\Rightarrow10\lambda+7=97$
$\Rightarrow10\lambda=90$
$\Rightarrow\lambda=9$
View full question & answer→Question 4265 Marks
Examine the differentiability of f, where f is defined by:
$\text{f(x)}=\begin{cases}1+\text{x},&\text{if x}\leq2\\5-\text{x},&\text{if x}>2\end{cases}$
at x = 2.
AnswerWe have, $\text{f(x)}=\begin{cases}1+\text{x},&\text{if x}\leq2\\5-\text{x},&\text{if x}>2\end{cases}$ at x = 2.
Let's first check continuity at x = 2
For continuity at x = 0,
$\text{L.H.L}=\lim\limits_{\text{x}\rightarrow2^-}(1+\text{x})=3$
$\text{R.H.L}=\lim\limits_{\text{x}\rightarrow2^+}(5-\text{x})=3$
$\text{f}(2)=1+2=3$
So, f(x) is continuous at x = 2
For differentiability at x = 2,
$\text{Lf}'(2)=\lim\limits_{\text{x}\rightarrow2^-}\frac{\text{f(x)}-\text{f}(2)}{\text{x}-2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(2-\text{h})-\text{f}(2)}{2-\text{h}-2}=\lim\limits_{\text{h}\rightarrow0}\frac{(1+2-\text{h})-3}{2-\text{h}-2}=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{h}}{-\text{h}}=1$
$\text{Rf}'(2)=\lim\limits_{\text{x}\rightarrow2^+}\frac{\text{f(x)}-\text{f}(2)}{\text{x}-2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(2+\text{h})-\text{f}(2)}{2+\text{h}-2}=\lim\limits_{\text{h}\rightarrow0}\frac{5-(2+\text{h})-3}{2+\text{h}-2}=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{h}}{\text{h}}=-1$
Since, Lf'(2) ≠ Rf'(2)
So, f(x) is not differentiable at x = 2.
View full question & answer→Question 4275 Marks
Extend the definition of the following by continuity $\text{f(x)}=\frac{1-\cos7(\text{x}-\pi)}{5(\text{x}-\pi)^2}$ at the point $\text{x}=\pi.$
AnswerGiven,
$\text{f(x)}=\frac{1-\cos7(\text{x}-\pi)}{5(\text{x}-\pi)^2},\text{ x}=\pi$
If f(x) is continuous at $\text{x}=\pi,$ then
$\lim_\limits{\text{x}\rightarrow\pi}\text{f(x)}=\text{f}(\pi)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow\pi}\frac{1-\cos7(\text{x}-\pi)}{5(\text{x}-\pi)^2}=\text{f}(\pi)$
$\Rightarrow\frac{2}{5}\lim_\limits{\text{x}\rightarrow\pi}\frac{\sin^2\Big(\frac{7(\text{x}-\pi)}{2}\Big)}{(\text{x}-\pi)^2}=\text{f}(\pi)$
$\Rightarrow\frac{2}{5}\times\frac{49}{4}\lim_\limits{\text{x}\rightarrow\pi}\frac{\sin^2\Big(\frac{7(\text{x}-\pi)}{2}\Big)}{\frac{49}{4}(\text{x}-\pi)^2}=\text{f}(\pi)$
$\Rightarrow\frac{2}{5}\times\frac{49}{4}\lim_\limits{\text{x}\rightarrow\pi}\frac{\sin^2\Big(\frac{7(\text{x}-\pi)}{2}\Big)}{\Big(\frac{7}{2}(\text{x}-\pi)\Big)^2}=\text{f}(\pi)$
$\Rightarrow\begin{bmatrix}\frac{2}{5}\times\frac{49}{4}\lim_\limits{\text{x}\rightarrow\pi}\frac{\sin\Big(\frac{7(\text{x}-\pi)}{2}\Big)}{\Big(\frac{7}{2}(\text{x}-\pi)\Big)}\end{bmatrix}^2=\text{f}(\pi)$
$\Rightarrow\frac{2}{5}\times\frac{49}{4}\times1=\text{f}(\pi)$
$\Rightarrow\frac{1}{5}\times\frac{49}{2}\times1=\text{f}(\pi)$
$\Rightarrow\frac{49}{10}=\text{f}(\pi)$
Hence, the given function will be continuous at $\text{x}=\pi,$ if $\text{f}(\pi)=\frac{49}{10}$
View full question & answer→Question 4285 Marks
A function f : R → R satisfies the equation f(x + y) = f(x)f(y) fot all x, y ∈ R, f(x) ≠ 0. Suppose that the function is differentiable at x = 0 and f'(0) = 2. Prove that f'(x) = 2 f(x).
AnswerLet f : R → R satisfies the equation $\text{f(x+y})=\text{f(x).f(y)},\forall\text{ x, y}\in\text{R, f(x)} \neq0.$Let f(x) is differentiable at x = 0 and f'(0) = 2.
$\Rightarrow\ \text{f}'(0)=\lim\limits_{\text{x}\rightarrow0}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$ $\Rightarrow\ 2=\lim\limits_{\text{x}\rightarrow0}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}$ $\Rightarrow\ 2=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(0+\text{h})-\text{f}(0)}{0+\text{h}}$ $\Rightarrow\ 2=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(0).\text{f(h)}-\text{f}(0)}{\text{h}}$ $\Rightarrow\ 2=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(0)[\text{f(h)-1]}}{\text{h}}$ $[\because\text{f}(0)=\text{f(h)}]\ \dots(\text{i})$ Also, $\text{f}'(\text{x})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x}+\text{h)}-\text{f(x)}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x)}.\text{f(h)}-\text{f(h)}}{\text{h}}$ $[\because\text{f(x+y)}=\text{f(x).f(y)}]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x)}[\text{f(h)}-1]}{\text{h}}=2$ $\text{f(x) }[\text{using Eq.(i)}]$ $\therefore\ \text{f}'(\text{x})=2\text{f(x)}$
View full question & answer→Question 4295 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\frac{\text{x}^4+\text{x}^3+2\text{x}^2}{\tan^{-1}\text{x}},&\text{if }\text{ x}\neq0\\10,&\text{if }\text{ x}=0\end{cases}$
AnswerWhen $\text{x}\neq0,$ we have $\text{f(x)}=\frac{\text{x}^4+\text{x}^3+2\text{x}^2}{\tan^{-1}\text{x}}$
We know that a polynomial is continuous for x < 0 and x > 0, Also the inverse trignometric function is continuous in its domain.
Here,$ X^4 + x^3 + 2x^2$ is polnomial, so is continuous for x < 0 and x > 0 and $\tan^{-1}\text{x}$ is also continuous for x < 0 and x > 0
So, the quotient function $\text{f(x)}=\frac{\text{x}^4+\text{x}^3+2\text{x}^2}{\tan^{-1}\text{x}}$ is continuous for each x < 0 and x > 0
Now, consider the point x = 0
$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(-\text{h})^4+(-\text{h})^3+2(-\text{h})^2}{\tan^{-1}(-\text{h})}=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^4-\text{h}^3+2\text{h}^2}{\tan^{-1}\text{h}}=0$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^4+\text{h}^3+2\text{h}^2}{\tan^{-1}\text{h}}=0$
$\text{f}(0)=10$
Thus, $\text{LHL}=\text{RHL}\neq\text{f}(0)$
Hence, the function is not continuous at x = 0
View full question & answer→Question 4305 Marks
If $\text{x}=\Big(\text{t}+\frac{1}{\text{t}}\Big)^\text{a},\text{y}=\text{a}^{\text{t}+\frac{1}{\text{t}}},$ find $\frac{\text{dy}}{\text{dx}}$
AnswerHere, $\text{x}=\Big(\text{t}+\frac{1}{\text{t}}\Big)^{\text{a}}$
Differentiating it with respect to t using chain rule,
$\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\left[\Big(\text{t}+\frac{1}{\text{t}}\Big)^{\text{a}}\right]$
$=\text{a}\Big(\text{t}+\frac{1}{\text{t}}\Big)^{\text{a-1}}\frac{\text{d}}{\text{dt}}\Big(\text{t}+\frac{1}{\text{t}}\Big)$
$\frac{\text{dx}}{\text{dt}}=\text{a}\Big(\text{t}+\frac{1}{\text{t}}\Big)^{\text{a-1}}\Big(\text{1}-\frac{1}{\text{t}^{2}}\Big)\ .....(\text{i})$
And, $\text{y}=\text{a}^{(\text{t}+\frac{1}{\text{t}})}$
Differentiating it with respect to t using chain rule,
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\bigg[\text{a}^{\big(\text{t}+\frac{1}{\text{t}}\big)}\bigg]$
$=\frac{\text{d}}{\text{dt}}\text{a}^{\big(\text{t}+\frac{1}{\text{t}}\big)}\times\log\frac{\text{d}}{\text{dt}}\left(\text{t}+\frac{1}{\text{t}}\right)$
$\frac{\text{dy}}{\text{dt}}=\text{a}^{\big(\text{t}+\frac{1}{\text{t}}\big)}\times\log\text{a}\Big(\text{t}-\frac{1}{\text{t}^{2}}\Big)\ .....(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{a}^{\big(\text{t}+\frac{1}{\text{t}}\big)}\times\log\text{a}\left(1-\frac{1}{\text{t}^{2}}\right)}{\text{a}\big(\text{t}+\frac{1}{\text{t}}\big)^{\text{a}-1}\big(1-\frac{1}{\text{t}}^{2}\big)}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{a}^{\big(\text{t}+\frac{1}{\text{t}}\big)}\times\log\text{a}}{\text{a}\big(\text{t}+\frac{1}{\text{t}}\big)^{\text{a}-1}}$
View full question & answer→Question 4315 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\frac{\text{e}^{\text{ax}}\sec\text{x}\times\log\text{x}}{\sqrt{1-2\text{x}}}$
AnswerHere,
$\text{y}=\frac{\text{e}^{\text{ax}}\sec\text{x}\times\log\text{x}}{\sqrt{1-2\text{x}}}\ .....(\text{i})$
$\Rightarrow\text{y}=\frac{\text{e}^{\text{ax}}\times\sec^\text{x}\times\log\text{x}}{(1-2\text{x})^\frac{1}{2}}$
Taking log on both the sides,
$\log\text{y}=\log\text{e}^{\text{ax}}+\log{\sec\text{x}}+\log\log\text{x}-\frac{1}{2}\log(1-2\text{x}) \\ \begin{bmatrix} \text{Since}, \log\Big(\frac{\text{A}}{\text{B}}\Big)=\log\text{A}-\log\text{B},\\ \log(\text{AB})=\log\text{A}+\log\text{B} \end{bmatrix}$
$\log\text{y}=\text{ax}+\log{\sec\text{x}}+\log\log\text{x}-\frac{1}{2}\log(1-2\text{x}) \\ \big[\text{Since}, \log\text{a}^\text{b}=\text{b}\log\text{a and }\log_\text{e}\text{e}=1\big]$
Differentiating it with respect to x using chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{ax})+\frac{\text{d}}{\text{dx}}(\log\sec\text{x})+\frac{\text{d}}{\text{dx}}(\log\log\text{x})-\frac{1}{2}\log(1-2\text{x})$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{a}+\frac{1}{\sec\text{x}}\frac{\text{d}}{\text{dx}}(\sec\text{x})+\frac{1}{\log\text{x}}\frac{\text{d}}{\text{dx}}-\frac{1}{2}\Big(\frac{1}{1-2\text{x}}\Big)\frac{\text{d}}{\text{dx}}(1-2\text{x})$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{a}+\frac{\sec\text{x}\tan\text{x}}{\sec\text{x}}+\frac{1}{(\log\text{x})}\Big(\frac{1}{\text{x}}\Big)-\frac{1}{2}\Big(\frac{1}{1-2\text{x}}\Big)(-2)$
$\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\text{a}+\tan\text{x}+\frac{1}{\text{x}\log\text{x}}+\frac{1}{1-2\text{x}}\Big]$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{ax}}\sec\text{x}\log\text{x}}{\sqrt{1-2\text{x}}}\Big[\text{a}+\tan\text{x}+\frac{1}{\text{x}\log\text{x}}+\frac{1}{1-2\text{x}}\Big]$
[Using equation (i)]
View full question & answer→Question 4325 Marks
If $x^{13} y^7 = (x + y)^{20}$, prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
AnswerHere,
$x^{13}y^7 = (x + y)^{20}$
Taking log on both the sides,
$\log(\text{x}^{13}\text{y}^7)=\log(\text{x}+\text{y})^{20}$
$13\log\text{x}+7\log\text{y}=20\log(\text{x}+\text{y})$
$\big[\text{Since},\log(\text{AB})=\log\text{A}+\log\text{B},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule,
$13\frac{\text{d}}{\text{dx}}(\log\text{x})+7\frac{\text{d}}{\text{dx}}(\log\text{y})=20\frac{\text{d}}{\text{dx}}\log(\text{x}+\text{y})$
$\frac{13}{\text{x}}+\frac{7}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{20}{\text{x}+\text{y}}\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})$
$\frac{13}{\text{x}}+\frac{7}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{20}{(\text{x}+\text{y})}\Big[1+\frac{\text{dy}}{\text{dx}}\Big]$
$\frac{7}{\text{y}}\frac{\text{dy}}{\text{dx}}-\frac{20}{(\text{x}+\text{y})}=\frac{20}{(\text{x}+\text{y})}-\frac{13}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{7}}{\text{y}}-\frac{20}{(\text{x}+\text{y})}\Big]=\frac{20}{(\text{x}+\text{y})}-\frac{13}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{2(\text{x}+\text{y})-20\text{y}}{\text{y}(\text{x}+\text{y})}\Big]=\Big[\frac{20\text{x}-13(\text{x}+\text{y})}{\text{x}(\text{x}+\text{y})}\Big]$
$\frac{\text{dy}}{\text{dx}}=\Big[\frac{20\text{x}-13\text{x}-13\text{y}}{\text{x}(\text{x}+\text{y})}\Big]\Big(\frac{\text{y}(\text{x}+\text{y})}{7\text{x}+7\text{y}-20\text{y}}\Big)$
$=\frac{\text{y}}{\text{x}}\Big(\frac{7\text{x}-13\text{y}}{7\text{x}-13\text{y}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
View full question & answer→Question 4335 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$\text{f}(\text{x})=\tan^{-1}\text{x}\text{ on }[0,1]$
AnswerWe have,$\text{f}(\text{x})=\tan^{-1}\text{x}$
Clearly, f(x) is continuous on 0, 1 and derivable on 0, 1
Thus, both the conditions of Lagrange's theorem are satisfied.
Concequently, there exist some $\text{c}\in-3,4$ such that
$\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}=\frac{\text{f}(1)-\text{f}(0)}{1}$
Now,
$\text{f}(\text{x})=\tan^{-1}\text{x}$
$\text{f}'(\text{x})=\frac{1}{1+\text{x}^2},\text{f}(1)=\frac{\pi}{4},\text{f}(0)=0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
$\Rightarrow\frac{1}{1+\text{x}^2}=\frac{\pi}{4}-0$
$\Rightarrow49\Big(\frac{\pi}{4}-1\Big)=\text{x}^2$
$\Rightarrow\text{x}=\pm\sqrt{\frac{4-\pi}{\pi}}$
Thus, $\text{c}=\sqrt{\frac{4-\pi}{\pi}}\in(0,1)$ such that $\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 4345 Marks
Find the values of a and b such that the function f defined by $\text{f(x)}=\begin{cases}\frac{\text{x}-4}{|\text{x}-4|}+\text{a},&\text{if x}<4\\\text{a+}\text{b},&\text{if x}=4\\\frac{\text{x}-4}{|\text{x}-4|}+\text{b},&\text{if x}>4\end{cases}$ is a continuous function at x = 4.
AnswerConsider, $\text{f(x)}=\begin{cases}\frac{\text{x}-4}{|\text{x}-4|}+\text{a},&\text{if x}<4\\\text{a+}\text{b},&\text{if x}=4\\\frac{\text{x}-4}{|\text{x}-4|}+\text{b},&\text{if x}>4\end{cases}$
At x = 4 $\text{L.H.L}=\lim\limits_{\text{x}\rightarrow4^-}\frac{\text{x}-4}{|\text{x}-4|}+\text{a}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{4-\text{h}-4}{|4-\text{h}-4|}+\text{a}=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{h}}{\text{h}}+\text{a}$
$=-1+\text{a}$
$\text{R.H.L}=\lim\limits_{\text{x}\rightarrow4^+}\frac{\text{x}-4}{|\text{x}-4|}+\text{b}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{4+\text{h}-4}{|4+\text{h}-4|}+\text{b}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\text{h}}+\text{b}=1+\text{b}$
$\text{f}(4)=\text{a}+\text{b}$
Since, f(x) is continuous at x = 0
$\therefore$ L.H.L = R.H.L = f(0)
⇒ -1 + a = 1 + b = a + b
⇒ -1 + a = a + b and 1 + b = a + b
$\therefore$ b = -1 and a = 1
View full question & answer→Question 4355 Marks
Find whether the function is differentiable at x = 1 and x = 2
$\text{f(x)}=\begin{cases}\text{x} & \text{x}\leq1\\2-\text{x} & 1\leq\text{x}\leq2\\-2+3\text{x}&\text{x}>2\end{cases}$
Answer$\text{f(x)}=\begin{cases}\text{x} & \text{x}\leq1\\2-\text{x} & 1\leq\text{x}\leq2\\-2+3\text{x}&\text{x}>2\end{cases}$
$\text{f(x)}=\begin{cases}\text{x} & \text{x}\leq1\\-1 & 1\leq\text{x}\leq2\\3-2\text{x}&\text{x}>2\end{cases}$
Now,
LHL $=\lim_\limits{\text{x}\rightarrow1^{-}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow1^{-}}1=1$
RHL $=\lim_\limits{\text{x}\rightarrow1^{+}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow1^{+}}-1=-1$
Since, at $\text{x}=1,\text{LHL}\neq\text{RHL}$
Hence, f(x) is not differentiable at x = 1
Again,
LHL $=\lim_\limits{\text{x}\rightarrow2^{-}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow2^{-}}-1=-1$
RHL $=\lim_\limits{\text{x}\rightarrow2^{+}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow2^{+}}3-2\text{x}=3-4=-1$
Since, at x = 2, LHL = RHL
Hence, f(x) is differentiable at x = 2.
View full question & answer→Question 4365 Marks
If $\tan^{-1}\Big(\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}\Big)=\text{a}$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\text{y}}{\text{x}}\frac{(1-\tan\text{a})}{(1+\tan\text{a})}$
AnswerWe have, $\tan^{-1}\Big(\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}\Big)=\text{a}$
$\Rightarrow\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}=\tan\text{a}$
$\Rightarrow\text{x}^2-\text{y}^2=\tan\text{a}(\text{x}^2+\text{y}^2)$
Differentiating with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}(\text{x}^2-\text{y}^2)=\tan\text{a}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{y}^2)$
$\Rightarrow\Big(2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}\Big)=\tan\text{a}\Big(2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}=2\text{x}\tan\text{a}+2\text{y}\tan\text{a}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow2\text{y}\tan\text{a}\frac{\text{dy}}{\text{dx}}+2\text{y}\frac{\text{dy}}{\text{dx}}=2\text{x}-2\text{x}\tan\text{a}$
$\Rightarrow2\text{y}(1+\tan\text{a})\frac{\text{dy}}{\text{dx}}=2\text{x}(1-\tan\text{a})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}\Big(\frac{1-\tan\text{a}}{1+\tan\text{a}}\Big)$
View full question & answer→Question 4375 Marks
Differentiate the following functions with respect to x:
$\text{e}^{\sin\text{x}}+(\tan\text{x})^\text{x}$
AnswerLet $\text{y}=\text{e}^{\sin\text{x}}+(\tan\text{x})^\text{x}$
$\Rightarrow\text{y}=\text{e}^{\sin\text{x}}+\text{e}^{\log(\tan\text{x})^\text{x}}$
$\Rightarrow\text{y}=\text{e}^{\sin\text{x}}+\text{e}^{\text{x}\log(\tan\text{x})}$
Differentiating with resepect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\sin\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big\{\text{e}^{\text{x}\log(\tan\text{x})}\big\}$
$=\text{e}^{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\text{e}^{\text{x}\log(\tan\text{x})}\frac{\text{d}}{\text{dx}}(\text{x}\log\tan\text{x})$
$=\text{e}^{\sin\text{x}}(\cos\text{x})+\text{e}^{\log(\tan\text{x})^\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\tan\text{x})+\log\tan\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$=\text{e}^{\sin\text{x}}(\cos\text{x})+(\tan\text{x})^\text{x}\Big[\frac{\text{x}}{\tan\text{x}}(\sec^2\text{x})+\log\tan\text{x}\Big]$
$=\text{e}^{\sin\text{x}}(\cos\text{x})+(\tan\text{x})^\text{x}\big[\text{x}\sec\text{x cosec x}+\log\tan\text{x}\big]$
View full question & answer→Question 4385 Marks
If $\text{y}=\cot^{-1}\Big\{\frac{\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}}{\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\theta}}\Big\},$ show that $\frac{\text{dy}}{\text{dx}}$ is independent of x.
AnswerLet $\text{y}=\cot^{-1}\Big[\frac{\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}}{\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\theta}}\Big] \ .....(\text{i})$
Then, $\frac{\sqrt{1+\sin\theta}+\sqrt{1-\sin\text{x}}}{\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\text{x}}}$
$=\frac{\big(\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}\big)^2}{\big(\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\text{x}}\big)\big(\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}\big)} $
$=\frac{(1+\sin\text{x})+(1-\sin\text{x})+2\sqrt{(1-\sin\text{x})(1+\sin\text{x})}}{(1+\sin\text{x})-(1-\sin\text{x})}$
$=\frac{2+2\sqrt{1-\sin^2\text{c}}}{2\sin\text{x}}$
$=\frac{1+\cos\text{x}}{\sin\text{x}}$
$=\frac{2\cos^2\frac{\text{x}}{2}}{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}$
$=\cot\frac{\text{x}}{2}$
Therefore, equation (i) becomes
$\text{y}=\cot^{-1}\Big(\cot\frac{\text{x}}{2}\Big)$
$\Rightarrow\ \text{y}=\frac{\text{x}}{2}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{1}{2}\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{1}{2}$
View full question & answer→Question 4395 Marks
If $\text{x}=\text{a}(\cos\theta+\theta\sin\theta),\text{y}=\text{a}(\sin\theta-\theta\cos\theta)$ prove that $\frac{\text{d}^2\text{x}}{\text{d}\theta^2}=\text{a}(\cos\theta-\theta\sin\theta),\frac{\text{d}^2}{\text{d}\theta^2}$ $=\text{a}(\sin\theta-\theta\cos\theta)\ \text{and}\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\sec^3\theta}{\text{a}\theta}$
AnswerIt is given that, $\text{x}=\text{a}(\cos\text{t}+\text{t}\sin\text{t}),\text{y}=\text{a}(\sin\text{t}-\text{t}\cos\text{t})$
$\therefore\frac{\text{dx}}{\text{dt}}=\text{a}.\frac{\text{d}}{\text{dt}}(\cos\text{t}+\text{t}\sin\text{t})$
$=\text{a}\Big[-\sin\text{t}+\sin\text{t}.\frac{\text{d}}{\text{dt}.}(\text{t})+\frac{\text{d}}{\text{dt}}(\sin\text{t})\Big]$
$=\text{a}[-\sin\text{t}+\sin\text{t}+t\cos\text{t}]=\text{at}\cos\text{t}$
$\frac{\text{dy}}{\text{dt}}=\text{a}.\frac{\text{d}}{\text{dt}}(\sin\text{t}-\text{t}\cos\text{t})$
$=\text{a}\Big[\cos\text{t}-\Big\{\cos\text{t}\frac{\text{d}}{\text{dt}}(\text{t})+\text{t}.\frac{\text{d}}{\text{dt}}(\cos\text{t})\Big\}\Big]$
$=\text{a}[\cos\text{t}-\{\cos\text{t}-\text{t}\sin\text{t}\}]=\text{at}\sin\text{t}$
then, $\therefore\frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{dt}}\Big)}{\Big(\frac{\text{dx}}{\text{dt}}\Big)}=\frac{\text{at}\sin\text{t}}{\text{at}\cos\text{t}}=\tan\text{t}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}(\tan\text{t})=\sec^2\text{t}.\frac{\text{dt}}{\text{dx}}$
$=\sec^2\text{t}.\frac{1}{\text{at}\cos\text{t}}\ \Big[\frac{\text{dx}}{\text{dt}}=\text{at}\cos\text{t}\Rightarrow\frac{\text{dt}}{\text{dx}}=\frac{1}{\text{at}\cos\text{t}}\Big]$
$=\frac{\sec^3\text{t}}{\text{at}},0<\text{t}<\frac{\pi}{2}$
View full question & answer→Question 4405 Marks
If $(\cos\text{x})^{\text{y}}=(\tan\text{y})^{\text{x}},$ Prove that $\frac{\text{dy}}{\text{dx}}=\frac{\log\tan\text{y}-\text{y}\tan\text{x}}{\log\cos\text{x}-\text{x}\sec\text{y cosec y}}$
AnswerHere,
$(\cos\text{x})^{\text{y}}=(\tan\text{y})^{\text{x}}$
Taking log on both sides,
$\log(\cos\text{x})^{\text{y}}=\log(\tan\text{y})^{\text{x}}$
$\text{y}\log(\cos\text{x})=\text{x}\log(\tan\text{y})$
$\big[\text{Since}, \log\text{e}^{\text{b}}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule and product rule,
$\frac{\text{d}}{\text{dx}}(\text{y}\log\cos\text{x})=\frac{\text{d}}{\text{dx}}(\text{x}\log\tan\text{y})$
$\Big(\text{y}\frac{\text{d}}{\text{dx}}\log\cos\text{x}+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}\Big) \\ =\Big(\text{x}\frac{\text{d}}{\text{dx}}\log\tan\text{y}+\log\tan\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big)$
$\Big(\text{y}\big(\frac{1}{\cos\text{x}}\big)\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}\Big) \\ =\Big(\text{x}\frac{1}{\tan\text{y}}\frac{\text{d}}{\text{dx}}(\tan\text{y})+\log\tan\text{y}(1)\Big)$
$\Big(\frac{\text{y}}{\cos\text{x}}(-\sin\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}\Big)\\ =\Big(\frac{\text{x}}{\tan\text{y}}(\sec^2\text{y})\Big)\frac{\text{dy}}{\text{dx}}+\log\tan\text{y}-\text{y}\tan\text{x}+\log\cos\text{x}\frac{\text{dy}}{\text{dx}} \\ =\Big(\sec\text{y cosec y}\times\text{y}\frac{\text{dy}}{\text{dx}}+\log\tan\text{y}\Big)$
$\frac{\text{dy}}{\text{dx}}\big[\log\cos\text{x}-\text{x}\sec\text{y cosec y}\big] \\ =\log\tan\text{y}+\text{y}\tan\text{x}$
$\frac{\text{dy}}{\text{dx}}=\Big[\frac{\log\tan\text{x}+\text{y}\tan\text{x}}{\log\cos\text{x}-\text{x}\sec\text{y cosec y}}\Big]$
View full question & answer→Question 4415 Marks
Verify Rolle's theorem for the following function on the indicated intervals $f(x) = x^2+ 5 x + 6 $on the interval $[-3, -2]$
AnswerHere, $f(x) = x^2+ 5 x + 6$ on $[-3, -2]$
f(x) is continuous is [-3, -2] and f(x) is differentiable is $(-3, -2)$ since it is a polynomial function.
Now,
$f(x) = x^2+ 5x + 6$
$f(-3) = (-3)^2+5(-3) + 6$
$= 9 - 15 + 6$
$f(-3) = 0 ....(i)$
$f(-2) = (-2)^2+ 5(-2) + 6$
$= 4 - 10 + 6$
$f(-2) = 20 ....(ii)$
From equation (i) and (ii),
$f(-3) = f(-2)$
So, Rolle's theorem is applicable is $[-3, -2]$, we have to show that
f'(c) = 0 as $\text{c}\in (-3,-2)$
Now,
$f(x) = x^2 + 5x + 6$
$f'(x) = 2x + 5$
$\Rightarrow f'(c) = 0$
$2c + 5 = 0$
$\text{c}=\frac{-5}{2}\in(-3,-2)$
So, Rolle's theorem is verified.
View full question & answer→Question 4425 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=(\tan\text{x})^{\log\text{x}}+\cos^2\big(\frac{\pi}{4}\big)$
AnswerHere,
$\text{y}=(\tan\text{x})^{\log\text{x}}+\cos^2\big(\frac{\pi}{4}\big)$
$\text{y}=\text{e}^{\log(\tan\text{x})^{\log\text{x}}}+\cos^2\big(\frac{\pi}{2}\big)$
$\text{y}=\text{e}^{\log\text{x}\log\tan\text{x}}+\cos\text{x}^2\big(\frac{\pi}{4}\big)$
$\big[\text{Since, e}^{\log\text{a}}=\text{a and}\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating ti using chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\log\text{x}\log\tan\text{x}})+\frac{\text{d}}{\text{dx}}\cos^2\big(\frac{\pi}{4}\big)$
$=\text{e}^{\log\text{x}\log\tan\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x}\log\tan\text{x})+0$
$=\text{e}^{\log(\tan\text{x})^{\log\text{x}}}\Big[\log\times\frac{\text{d}}{\text{dx}}(\log\tan\text{x})+\log\tan\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})\Big]$
$=(\tan\text{x})^{\log\text{x}}\Big[\log\times\Big(\frac{1}{\tan\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\tan\text{x})+\log\tan\text{x}\Big(\frac{1}{\text{x}}\Big)\Big]$
$=(\tan\text{x})^{\log\text{x}}\Big[\log\times\Big(\frac{1}{\tan\text{x}}\Big)\big(\sec^2\text{x}\big)+\frac{\log\tan\text{x}}{\text{x}}\Big]$
$\frac{\text{dy}}{\text{dx}}=(\tan\text{x})^{\log\text{x}}\Big[\log\times\Big(\frac{\sec^2\text{x}}{\tan\text{x}}\Big)+\frac{\log\tan\text{x}}{\text{x}}\Big]$
View full question & answer→Question 4435 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\text{a}+\text{b}\tan\text{x}}{\text{b}-\text{a}\tan\text{x}}\Big)$
AnswerLet, $\text{y}=\tan^{-1}\Big[\frac{\text{a}+\text{b}\tan\text{x}}{\text{b}-\text{a}\tan\text{x}}\Big]$
$\Rightarrow\text{y}=\tan^{-1}\bigg[\frac{\frac{\text{a}+\text{b}\tan\text{x}}{\text{b}}}{\frac{\text{b}-\text{a}\tan\text{x}}{\text{b}}}\bigg]$
$\Rightarrow\text{y}=\tan^{-1}\bigg[\frac{\frac{\text{a}}{\text{b}}+\tan\text{x}}{1-\frac{\text{a}}{\text{b}}\tan\text{x}}\bigg]$
$\Rightarrow\text{y}=\tan^{-1}\Bigg[\frac{\tan\Big(\tan^{-1}\frac{\text{a}}{\text{b}}\Big)+\tan\text{x}}{1-\tan\Big(\tan^{-1}\frac{\text{a}}{\text{b}}\Big)\times\tan\text{x}}\Bigg]$
$\Rightarrow\text{y}=\tan^{-1}\Big[\tan\Big(\tan^{-1}\frac{\text{a}}{\text{b}}+\text{x}\Big)\Big]$
$\Rightarrow\text{y}=\tan^{-1}\big(\frac{\text{a}}{\text{b}}\big)+\text{x}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0+1$
$\therefore\ \frac{\text{dy}}{\text{dx}}=1$
View full question & answer→Question 4445 Marks
Differentiate the following functions with respect to x:
$(\sin\text{x})^{\log\text{x}}$
AnswerLet $\text{y}=(\sin\text{x})^{\log\text{x}}\ .....(\text{i})$
Taking log on both sides,
$\log\text{y}=\log(\sin\text{x})^{\log\text{x}}$
$\Rightarrow\log\text{y}=\log\text{x}\log\sin\text{x}$
Differentiating with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log\text{x}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log\text{x}\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\Big(\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\log\text{x}}{\sin\text{x}}(\cos\text{x})+\frac{\log\sin\text{x}}{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\log\text{x}\cot\text{x}+\frac{\log\sin\text{x}}{\text{x}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(\sin\text{x})^{\log\text{x}}\Big[\log\text{x}\cot\text{x}+\frac{\log\sin\text{x}}{\text{x}}\Big]$
[Using equation (i)]
View full question & answer→Question 4455 Marks
If $\text{xy}\log(\text{x}+\text{y})=1,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\text{x}^2\text{y}+\text{x}+\text{y})}{\text{x}(\text{xy}^2+\text{x}+\text{y})}$
AnswerHere,
$\text{xy}\log(\text{x}+\text{y})=1\ .....(\text{i})$
Differentiaitng with respect to x using chain rula, product rule,
$\frac{\text{dy}}{\text{dx}}(\text{xy}\log(\text{x}+\text{y}))=\frac{\text{d}}{\text{dx}}(1)$
$\text{xy}\frac{\text{d}}{\text{dx}}\log(\text{x}+\text{y})+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}\log(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})=0$
$\frac{\text{xy}}{(\text{x}+\text{y})}\Big(1+\frac{\text{dy}}{\text{dx}}\Big)+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}\log(\text{x}+\text{y})(1)=0$
$\Big(\frac{\text{xy}}{\text{x}+\text{y}}\Big)\Big(1+\frac{\text{dy}}{\text{dx}}\Big)+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{x}\log(\text{x}+\text{y})=0$
$\Big(\frac{\text{xy}}{\text{x}+\text{y}}\Big)\frac{\text{dy}}{\text{dx}}+\frac{\text{xy}}{\text{x}+\text{y}}+\text{x}\Big(\frac{1}{\text{xy}}\Big)\frac{\text{dy}}{\text{dx}}+\text{y}\Big(\frac{1}{\text{xy}}\Big)=0$
[Using equation (i)]
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{xy}}{\text{x}+\text{y}}+\frac{1}{\text{y}}\Big]=-\Big[\frac{1}{\text{x}}+\frac{\text{xy}}{\text{x}+\text{y}}\Big]$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{xy}^2+\text{x}+\text{y}}{(\text{x}+\text{y})\text{y}}\Big]=-\Big[\frac{\text{x}+\text{y}+\text{x}^2\text{y}}{\text{x}(\text{x}+\text{y})}\Big]$
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}\Big(\frac{\text{x}+\text{y}+\text{x}^2\text{y}}{\text{x}+\text{y}+\text{xy}^2}\Big)$
View full question & answer→Question 4465 Marks
Differentiate the following functions with respect to x:
$(\log\text{x})^\text{x}$
AnswerLet $\text{y}=(\log\text{x})^\text{x}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(\log\text{x})^\text{x}$
$\Rightarrow\log\text{y}=\text{x}\log(\log\text{x})\ \big[\text{Since}, \log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating with respect to x, using product rule, chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\log(\log\text{x})+\log\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})$
$=\text{x}\frac{1}{\log\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\log\text{x}(1)$
$=\frac{\text{x}}{\log\text{x}}\Big(\frac{1}{\text{x}}\Big)+\log\log\text{x}$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\log\text{x}}+\log\log\text{x}$
$\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{1}{\log\text{x}}+\log\log\text{x}\Big]$
$\frac{\text{dy}}{\text{dx}}=(\log\text{x})^\text{x}\Big[\frac{1}{\log\text{x}}+\log\log\text{x}\Big]$
[Using equation (i)]
View full question & answer→Question 4475 Marks
If $(\sin\text{x})^{\text{y}}=\text{x}+\text{y},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{1-(\text{x}+\text{y})\text{y}\cot\text{x}}{(\text{x}+\text{y})\log\sin\text{x}-1}$
AnswerHere,
$(\sin\text{x})^{\text{y}}=\text{x}+\text{y}$
Taking log on both the sides,
$\log(\sin\text{x})^\text{y}=\log(\text{x}+\text{y})$
$\text{y}\log(\sin\text{x})=\log(\text{x}+\text{y})\ \big[\text{Since},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule, product rule,
$\frac{\text{d}}{\text{dx}}(\text{y}\log(\sin\text{x}))=\frac{\text{d}}{\text{dx}}\log(\text{x}+\text{y})$
$\text{y}\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\log\sin\text{x}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}+\text{y}}\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})$
$\frac{\text{y}}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}\Big[1+\frac{\text{dy}}{\text{dx}}\Big]$
$\frac{\text{y}(\cos\text{x})}{(\sin\text{x})}+\log\sin\text{x}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}+\frac{1}{(\text{x}+\text{y})}\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}\Big(\log\sin\text{x}-\frac{1}{\text{x}+\text{y}}\Big)=\frac{1}{(\text{x}+\text{y})}-\text{y}\cot\text{x}$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{(\text{x}+\text{y})\log\sin\text{x}-1}{(\text{x}+\text{y})}\Big)=\Big(\frac{1-\text{y}(\text{x}+\text{y})\cot\text{x}}{\text{x}+\text{y}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\Big(\frac{1-\text{y}(\text{x}+\text{y})\cot\text{x}}{(\text{x}+\text{y})\log\sin\text{x}-1}\Big)$
View full question & answer→Question 4485 Marks
If $\text{y}=(\sin\text{x}-\cos\text{x})^{\sin\text{x}-\cos\text{x}},\frac{\pi}{4}<\text{x}<\frac{3\pi}{4},$ find $\frac{\text{dy}}{\text{dx}}$
AnswerWe have, $\text{y}=(\sin\text{x}-\cos\text{x})^{\sin\text{x}-\cos\text{x}}\ .....(\text{i})$
Taking log on both sides,
$\log\text{y}=\log(\sin\text{x}-\cos\text{x})^{\sin\text{x}-\cos\text{x}}$
$\Rightarrow\log\text{y}=(\sin\text{x}-\cos\text{x})^{\sin\text{x}-\cos\text{x}}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log(\sin\text{x}-\cos\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x}-\cos\text{x}) \\ +(\sin\text{x}-\cos\text{x})\frac{\text{d}}{\text{dx}}\log(\sin\text{x}-\cos\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=(\sin\text{x}-\cos\text{x})(\sin\text{x}-\cos\text{x}) \\ +\frac{(\sin\text{x}-\cos\text{x})}{(\sin\text{x}-\cos\text{x})}\frac{\text{d}}{\text{dx}}(\sin\text{x}-\cos\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=(\cos\text{x}+\sin\text{x})\log(\sin\text{x}-\cos\text{x})+(\cos\text{x}+\sin\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=(\cos\text{x}+\sin\text{x})[1+\log(\sin\text{x}-\cos\text{x})]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\big[(\cos\text{x}+\sin\text{x})\big\{1+\log(\sin\text{x}-\cos\text{x})\big\}\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(\sin\text{x}-\cos\text{x})^{(\sin\text{x}-\cos\text{x})} \\ \big[(\cos\text{x}+\sin\text{x})\big\{1+\log(\sin\text{x}-\cos\text{x})\big\}\big]$
View full question & answer→Question 4495 Marks
Differentiate the following functions with respect to x:
$\frac{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}$
AnswerLet, $\text{y}=\frac{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\frac{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}\Big]$
$=\Bigg[\frac{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)\frac{\text{d}}{\text{dx}}\big(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}\big)-\big(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}\big)\frac{\text{d}}{\text{dx}}\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)}{\big(\text{e}^{2\text{x}}-\text{e}^{4\text{x}}\big)^2}\Bigg]$
[Using quotient rule and chain rule]
$=\frac{(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}})\Big[\text{e}^{2\text{x}}\frac{\text{d}}{\text{dx}}(2\text{x})+\text{e}^{-2\text{x}}\frac{\text{d}}{\text{dx}}(-2\text{x})\Big]-\big(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}\big)\Big[\text{e}^{2\text{x}}\frac{\text{d}}{\text{dx}}(2\text{x})-\text{e}^{-2\text{x}}\frac{\text{d}}{\text{dx}}(-2\text{x})\Big]}{\big(\text{e}^{2\text{x}}-\text{e}^{2\text{x}}\big)^2}$
$=\frac{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)\big(2\text{e}^{2\text{x}}-2\text{e}^{-2\text{e}}\big)-\big(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}\big)\big(2\text{e}^{2\text{e}}+2\text{e}^{-2\text{x}}\big)}{\big(\text{e}^{2\text{e}}-\text{e}^{-2\text{x}}\big)^2}$
$=\frac{2\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)^2-2\big(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}\big)^2}{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)^2}$
$=\frac{2\big[\text{e}^{4\text{x}}+\text{e}^{-4\text{x}}-2\text{e}^{2\text{x}}\text{e}^{-2\text{x}}-\text{e}^{4\text{x}}-\text{e}^{-4\text{x}}-2\text{e}^{2\text{x}}\text{e}^{-2\text{x}}\big]}{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)^2}$
$=\frac{-8}{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)^2}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}\Big)=\frac{-8}{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)^2}$
View full question & answer→Question 4505 Marks
If $\text{x}=\text{e}^{\cos2\text{t}}$ and $\text{y}=\text{e}^{\sin2\text{t}},$ prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}\log\text{x}}{\text{x}\log\text{y}}$
AnswerWe have, $\text{x}=(\text{e}^{\cos2\text{t}})$ and $\text{y}=\text{e}^{\sin2\text{t}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\big(\text{e}^{\cos2\text{t}}\big)$ and $\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\big(\text{e}^{\sin2\text{t}}\big)$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\text{e}^{\cos2\text{t}}\frac{\text{d}}{\text{dt}}(\cos2\text{t})$ and $\frac{\text{dx}}{\text{dt}}=\text{e}^{\cos2\text{t}}\frac{\text{d}}{\text{dt}}(\cos2\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\text{e}^{\cos2\text{t}}(-\sin2\text{t})\frac{\text{d}}{\text{dt}}(2\text{t})$ and $\frac{\text{dy}}{\text{dt}}=\text{e}^{\sin2\text{t}}(\cos2\text{t})\frac{\text{d}}{\text{dt}}(2\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=-2\sin 2\text{t}\text{e}^{\cos2\text{t}}$ and $\frac{\text{dy}}{\text{dt}}=2\cos 2\text{t}\text{e}^{\sin2\text{t}}$
$\because\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{2\cos2\text{te}^{\sin2\text{t}}}{-2\sin2\text{te}^{\cos2\text{t}}}$
$\begin{bmatrix} \because\text{x}=\text{e}^{\cos2\text{t}}\Rightarrow\log\text{x}=\cos2\text{t} \\ \text{y}=\text{e}^{\sin2\text{t}}\Rightarrow\log\text{y}=\sin2\text{t} \end{bmatrix}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}\log\text{x}}{\text{x}\log\text{y}}$
View full question & answer→