Question 12 Marks
If f'(1) = 2 and $\text{y}=\text{f}(\log_\text{e}\text{x}),$ find $\frac{\text{d}}{\text{dx}}\text{at x}=\text{e}.$
AnswerWe have, f'(1) = 2 and $\text{y}=\text{f}(\log_\text{e}\text{x})$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\text{f}'(\log_\text{e}\text{x})\times\frac{\text{d}}{\text{dx}}(\log_\text{e}\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{f}'(\log_\text{e}\text{x})\big(\frac{1}{\text{x}}\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{f}'(\log_\text{e}\text{e})\big(\frac{1}{\text{e}}\big) \big[\because\text{x}=\text{e}\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{f}'(1)\big(\frac{1}{\text{e}}\big) \big[\because\log_\text{e}\text{e}=1\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2}{\text{e}} \big[\because\text{f}'(1)=2\big]$
View full question & answer→Question 22 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{b}\sin^2\theta\text{ and y}=\text{a}\cos^2\theta$
AnswerHere $\text{x}=\text{b}\sin^2\theta\text{ and y}=\text{a}\cos^2\theta$Then,
$\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}(\text{b}\sin^2\theta)=2\text{b}\sin\theta\cos\theta$
$\frac{\text{dy}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}(\text{a}\cos^2\theta)=2\text{b}\cos\theta\sin\theta$
$\therefore\frac{\text{dy}}{\text{d}\text{x}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{-2\text{a}\cos\theta\sin\theta}{2\text{b}\sin\theta\cos\theta}=\frac{-\text{a}}{\text{b}}$
View full question & answer→Question 32 Marks
If y = x|x|, find $\frac{\text{dy}}{\text{dx}}\text{ for x 0}<0.$
AnswerWe have, y = x|x|
$\Rightarrow\text{y}=\text{x}(-\text{x})\big(\because\text{x}>0\big)$
$\Rightarrow\text{y}=-\text{x}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(-\text{x}^2)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-2\text{x}$
View full question & answer→Question 42 Marks
If $-\frac{\pi}{2}<\text{x}<0\text{ and y}=\tan^{-1}\sqrt{\frac{1-\cos2\text{x}}{1+\cos2\text{x}}},$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerWe have, $\text{y}=\tan^{-1}\sqrt{\frac{1-\cos2\text{x}}{1+\cos2\text{x}}}$
$\Rightarrow\text{y}=\tan^{-1}\sqrt{\frac{2\sin^2\text{x}}{2\cos^2\text{x}}}$
$\Rightarrow\text{y}=\tan^{-1}\sqrt{\tan^2\text{x}}$
$\Rightarrow\text{y}=\tan^{-1}(\tan\text{x})$
$\Big[\because\tan^{-1}(\tan\text{x})=-\text{x},\text{if x}\in\big(-\frac{\pi}{2},0\big)\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-1$
View full question & answer→Question 52 Marks
If $\text{y}=\sec^{-1}\Big(\frac{\text{x}+1}{\text{x}-1}\Big)+\sin^{-1}\Big(\frac{\text{x}-1}{\text{x}+1}\Big),$ then write the value of $\frac{\text{dy}}{\text{dx}}.$
AnswerWe have, $\text{y}=\sec^{-1}\Big(\frac{\text{x}+1}{\text{x}-1}\Big)+\sin^{-1}\Big(\frac{\text{x}-1}{\text{x}+1}\Big)$
$\Rightarrow\text{y}=\cos^{-1}\Big(\frac{\text{x}-1}{\text{x}+1}\Big)+\sin^{-1}\Big(\frac{\text{x}-1}{\text{x}+1}\Big)$
$\Big[\because\sec^{-1}\text{x}=\cos^{-1}\big(\frac{1}{\text{x}}\big)\Big]$
$\Rightarrow\text{y}=\frac{\pi}{2}\Big[\because\sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=0$
View full question & answer→Question 62 Marks
Differentiate $x^2$ with respect to $x^3$.
AnswerLet $u = x^2$ and $v = x^3$
$\Rightarrow\frac{\text{du}}{\text{dx}}=2\text{x and }\frac{\text{dv}}{\text{dx}}=3\text{x}^2$
$\therefore\frac{\text{du}}{\text{dv}}=\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{2\text{x}}{3\text{x}^2}=\frac{2}{3\text{x}}$
View full question & answer→Question 72 Marks
If $\text{f}'\text{(x)}=\sqrt{2\text{x}^2-1}$ and $\text{y}=\text{f}(\text{x}^2),$ then find $\frac{\text{dy}}{\text{dx}}\text{at x}=1.$
AnswerHere,
$\text{f}'\text{(x)}=\sqrt{2\text{x}^2-1}$
and $\text{y}=\text{f}\big(\text{x}^2\big)$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\text{f}\big(\text{x}^2\big)$
$=\text{f}'\big(\text{x}^2\big)\frac{\text{d}}{\text{dx}}\big(\text{x}^2\big)$
$=\text{d}'\big(\text{x}^2\big)\times2\text{x}$
$\frac{\text{dy}}{\text{dx}}=2\text{xf}'\big(\text{x}^2\big)$
Put x = 1
$\frac{\text{dy}}{\text{dx}}=2(1)\text{f}'(1)$
$=2\times\text{f}'(1)$
$\frac{\text{dy}}{\text{dx}}=2\times1$
$\big[\text{Since},\text{f}'(1)=\sqrt{2(1)^2-1}=\sqrt{2-1}=1\big]$
$\frac{\text{dy}}{\text{dx}}=2$
View full question & answer→Question 82 Marks
If $\pi\leq\text{x}\leq3\pi$ and $\text{y}=\cos^{-1}(\cos\text{x}),$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerWe have,
$\text{y}=\cos^{-1}(\cos\text{x})$
$\Rightarrow\text{y}=2\pi-\text{x}$
$\big[\because\cos^{-1}(\cos\text{x})=2\pi-\text{x},\text{if x}\in[\pi,2{\pi}]\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(2\pi-\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=0-1$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-1$
View full question & answer→Question 92 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$x = at^2$ and $y = 2$ at
AnswerWe have $x = at^2$ and $y = 2$ at
$\Rightarrow\frac{\text{dx}}{\text{dx}}=2\text{at}\text{ and }\frac{\text{dy}}{\text{dx}}=2\text{a}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{2\text{a}}{2\text{at}}=\frac{1}{\text{t}}$
View full question & answer→Question 102 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{a}\cos\theta$ and $\text{y}=\text{b}\sin\theta$
AnswerWe have, $\text{x}=\text{a}\cos \theta$ and $\text{y}=\text{b}\cos\theta$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=-\text{a}\sin\theta$ and $\frac{\text{dy}}{d\theta}=\text{b}\cos\theta$
$\therefore\frac{\text{dy}}{{\text{d}}\theta}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{\text{b}\cos\theta}{-\text{a}\sin\theta}=\frac{-\text{b}}{\text{a}}\cot\theta$
View full question & answer→Question 112 Marks
If $\text{y}=\sin^{-1}\text{x}+\cos^{-1}\text{x},$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerWe have, $\text{y}=\sin^{-1}\text{x}+\cos^{-1}\text{x}$
$\Rightarrow\text{y}=\frac{\pi}{2}$
$\Big[\because\sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=0$
View full question & answer→Question 122 Marks
If $\text{y}=\sin^{-1}(\sin\text{x}),-\frac{\pi}{2}\leq\text{x}\leq\frac{\pi}{2}.$ Then, wrrite tha value of $\frac{\text{dy}}{\text{dx}}\text{ for x}\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big).$
AnswerWe have, $\text{y}=\sin^{-1}(\sin\text{x})$
$\Rightarrow\text{y}=\text{x}$
$\Big[\because\sin^{-1}(\sin\text{x})=\text{x},\text{ if x}\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1$
View full question & answer→Question 132 Marks
If $\text{y}=\sin^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big),$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere,
$\text{y}=\sin^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$\Big[\text{Since},\sin^{-1}\text{x}+\cos^{-1}=\frac{\pi}{2}\Big]$
So,
$\text{y}=\frac{\pi}{2}$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0$
View full question & answer→Question 142 Marks
If |x| < 1 and $\text{y}=1+\text{x}+\text{x}^2+\ .....\ \text{to}\ \infty,$ then find the value of $\frac{\text{dy}}{\text{dx}}.$
AnswerWe have, $\text{y}=1+\text{x}+\text{x}^2+\ .....\ \text{to}\ \infty,$
$\Rightarrow\text{y}=\frac{1}{1-\text{x}}$
[$\because$ It is a G.P with first term 1 and common ration x]
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{1}{1-\text{x}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{1}{(1-\text{x})^2}\frac{\text{d}}{\text{dx}}(1-\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{1}{(1-\text{x})^2}(-1)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{1}{(1-\text{x})^2}$
View full question & answer→Question 152 Marks
If $\frac{\pi}{2}\leq\text{x}\leq\frac{3\pi}{2}$ and $\text{y}=\sin^{-1}(\sin\text{x}),$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere,
$\text{y}=\sin^{-1}(\sin\text{x}),\text{x}\in\Big[\frac{\pi}{2},\frac{3\pi}{1}\Big]$
$\Big[\text{Since},\sin^{-1}(\sin\text{x})=\text{x},\text{if x}\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\pi-\text{x}$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\pi-\text{x})$
$0-1$
$\frac{\text{dy}}{\text{dx}}=-1$
View full question & answer→Question 162 Marks
If $\text{f(x)}=\log_\text{e}(\log_\text{e}\text{x}),$ then write the value of f'(e).
AnswerWe have, $\text{f(x)}=\log_\text{e}(\log_\text{e}\text{x})$
Differentiating with respect to x,
$\text{f}'\text{(x)}=\frac{1}{\log_\text{e}\text{x}}\frac{\text{d}}{\text{dx}}(\log_\text{e}\text{x})$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{\log_\text{e}\text{x}}\Big(\frac{1}{\text{x}}\Big)$
$\Rightarrow\text{f}'\text{(e)}=\frac{1}{\log_\text{e}\text{e}}\Big(\frac{1}{\text{e}}\Big) \big[\because\text{x}=\text{e}\big]$
$\Rightarrow\text{f}'\text{(e)}=\frac{1}{\text{x}}\big[\because\log_\text{e}=1\big]$
View full question & answer→Question 172 Marks
if f(1) = 4, f'(1) = 2, find the value of the derivative of $\log\Big(\text{f}\big(\text{e}^\text{x}\big)\Big)$ w.r.t x at the point x = 0.
AnswerWe have, f(1) = 4 and f'(1) = 2
Let $\text{y}=\log\big\{\text{f}\big(\text{e}^\text{x}\big)\big\}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\log\big\{\text{f}\big(\text{e}^\text{x}\big)\big\}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{f}\big(\text{e}^\text{x}\big)}\times\frac{\text{d}}{\text{dx}}\big\{\text{f}\big(\text{e}^\text{x}\big)\big\}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{f}(\text{e}^\text{x})}\times\text{f}'\big(\text{e}^\text{x}\big)\times\frac{\text{d}}{\text{dx}}\big(\text{e}^\text{x}\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}\text{f}'\big(\text{e}^\text{x}\big)}{\text{f}\big(\text{e}^\text{x}\big)}$
Putting x = 0, we get,
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^0\text{f}'\big(\text{e}^0\big)}{\text{f}\big(\text{e}^0\big)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1\text{f}'(1)}{\text{f}(1)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2}{4}\big[\because\text{f}'(1)=2\text{ and f}(1)=4\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2}$
View full question & answer→Question 182 Marks
Differentiate the following with respect to x:
$\cos^{-1}(\sin\text{ x})$
AnswerLet $\text{f(x)}=\cos^{-1}(\sin\text{x})$
We observe that this function is defined for all real numbers.
$\text{f(x)}=\cos^{-1}(\sin\text{x})$
$=\cos^{-1}\Big[\cos\Big(\frac{\pi}{2}-\text{x}\Big)\Big]=\frac{\pi}{2}-\text{x}$
Thus, $\text{f(x)}=\frac{\text{d}}{\text{dx}}\Big(\frac{\pi}{2}-\text{x}\Big)=-1$
View full question & answer→Question 192 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{a}(1-\cos\theta)\text{ and y}=\text{a}(\theta+\sin\theta)\text{ at }\theta=\frac{\pi}{1}$
AnswerWe have, $\text{x}=\text{a}(1-\cos\theta)\text{ and y}=\text{a}(\theta+\sin\theta)$
$\therefore\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}[\text{a}(1-\cos\theta)]=\text{a}(\sin\theta)$
and
$\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}[\text{a}(\theta+\sin\theta)]=\text{a}(1+\cos\theta)$
$\therefore\Big[\frac{\text{dy}}{\text{dx}}\Big]_{\theta=\frac{\pi}{2}}=\bigg[\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}\bigg]_{\theta=\frac{\pi}{2}} \\ =\Big[\frac{\text{a}(1+\cos\theta)}{\text{a}(\sin\theta)}\Big]_{\theta=\frac{\pi}{2}}=\frac{\text{a}(1+0)}{\text{a}}=1$
View full question & answer→Question 202 Marks
If f(x) = x + 1 then write the value of $\frac{\text{d}}{\text{dx}}\text{ fof }\text{(x)}.$
AnswerHere,
f(x) = x + 1
(fof)(x) = f(f(x))
= f(x + 1)
= (x + 1) + 1
(fof)(x) = x + 2
Differentiating it with respect to x,
$\frac{\text{d}}{\text{dx}}(\text{fof})\text{(x)}=\frac{\text{d}}{\text{dx}}\text{(x)}+\frac{\text{d}}{\text{dx}}(2)$
$=1+0$
$\frac{\text{d}}{\text{dx}}\text{(fof)}\text{(x)}=1$
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