3 Marks Question

Question 513 Marks

If $\text{y}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big),$ write the value of $\frac{\text{dy}}{\text{dx}}\text{ for x}>1.$

Answer

We have, $\text{y}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
Putting $\text{x}=\tan\theta$
$\Rightarrow 1 <\tan\theta<\infty$
$\Rightarrow\frac{\pi}{4}<\theta<\frac{\pi}{2}$
$\frac{\pi}{2}<2\theta<\pi$
$\therefore\text{y}=\sin^{-1}(\sin2\theta)$
$\Rightarrow\text{y}=\sin^{-1}\big\{\sin(\pi-2\theta)\big\}$
$\Rightarrow\text{y}=\pi-2\theta$
$\Rightarrow\text{y}=\pi-2\tan^{-1}\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=0-\frac{2}{1+\text{x}^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{2}{1+\text{x}^2}$

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Question 523 Marks

Differentiate the following functions with respect to x:
$\sin(3\text{x}+5)$

Answer

Consider $\text{y}=\sin(3\text{x}+5)$
Differentiate y with the respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\sin(3\text{x}+5)\big)$
$=\cos(3\text{x}+5)\frac{\text{d}}{\text{dx}}(3\text{x}+5)$
[using chain rule]
$=\cos(3\text{x}+5)\times[3(1)+0]$
$=3\cos(3\text{x}+5)$
Hence, the solution is $\frac{\text{d}}{\text{dx}}(\sin(3\text{x}+5))=3\cos(3\text{x}+5)$

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Question 533 Marks

If $\text{x}=\text{a}(\theta-\sin\theta)\text{ and},\text{y}=\text{a}(1+\cos\theta),$ find $\frac{\text{dy}}{\text{dx}}\text{ at }\theta=\frac{\pi}{3}$

Answer

Here,
$\text{x}=\text{a}(\theta-\sin\theta)\text{ and y}=\text{a}(1+\cos\theta)$
Then,
$\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\big[\text{a}(\theta-\sin\theta)\big]=\text{a}(1-\cos\theta)$
$\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\big[\text{a}(1+\sin\theta)\big]=\text{a}(1-\sin\theta)$
$\therefore\frac{\text{dy}}{\text{dx}}=\Bigg[\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{-\text{a}\sin\theta}{\text{a}(1-\cos\theta)}\Bigg]_{\theta=\frac{\pi}{3}}$
$=-\frac{\sin\frac{\pi}{2}}{1-\cos\frac{\pi}{3}}=\frac{\frac{\sqrt{3}}{2}}{1-\frac{1}{2}}=-\sqrt{3}$

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Question 543 Marks

If $\text{y}=\sqrt{\log\text{x}+\sqrt{\log\text{x}+\sqrt{\log\text{x}+\ .... \text{to }\infty}}},$ prove that $(2\text{y}-1)\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$

Answer

We have, $\text{y}=\sqrt{\log\text{x}+\sqrt{\log\text{x}+\sqrt{\log\text{x}+\ .... \text{to }\infty}}}$
$\Rightarrow\text{y}=\sqrt{\log\text{x}+\text{y}}$
Squaring both sides, we get,
$\text{y}^2=\log\text{x}+\text{y}$
$=2\text{y}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}+\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(2\text{y}-1)=\frac{1}{\text{x}}$

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Question 553 Marks

Differentiate the following functions with respect to x:
$(\log\sin\text{x})^2$

Answer

Let $\text{y}=(\log\sin\text{x})^2$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\log\sin\text{x})^2$
$=2(\log\sin\text{x})\frac{\text{d}}{\text{dx}}(\log\sin\text{x})$
$=2(\log\sin\text{x})\times\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})$
$=2(\log\sin\text{x})\times\frac{1}{\sin\text{x}}\cos\text{x}$
$=2(\log\sin\text{x})\cot\text{x}$
So,
$\frac{\text{d}}{\text{dx}}(\log\sin\text{x})^2=2(\log\sin\text{x})\cot\text{x}$

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Question 563 Marks

Differentiate the following functions with respect to x:
$\sin(\log\sin\text{x})$

Answer

Consider $\text{y}=\sin(\log\sin\text{x})$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sin(\log\sin\text{x})$
$=\cos(\log\sin\text{x})\frac{\text{d}}{\text{dx}}(\log\sin\text{x})$
[Using chain rule]
$=\cos(\log\sin\text{x})\times\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}0\sin\text{x}$
$=\cos(\log\sin\text{x})\frac{\cos\text{x}}{\sin\text{x}}$
$=\cos(\log\sin\text{x})\times\cot\text{x}$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}(\sin(\log\sin\text{x}))=\cos(\log\sin\text{x})\text{x}\cot\text{x}$

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Question 573 Marks

Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\text{a}+\text{x}}{1-\text{ax}}\Big)$

Answer

Let $\text{y}=\tan^{-1}\Big(\frac{\text{a}+\text{x}}{1-\text{ax}}\Big)$
$\text{y}=\tan^{-1}\text{a}+\tan^{-1}\text{x}$
$\Big[\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\tan^{-1}\text{a})+\frac{\text{d}}{\text{dx}}(\tan^{-1}\text{x})$
$=0+\frac{1}{1+\text{x}^2}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\text{x}^2}$

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Question 583 Marks

Differentiate the following functions with respect to x:
$\text{e}^{\sin\sqrt{\text{x}}}$

Answer

Let, $\text{y}=\text{e}^{\sin\sqrt{\text{x}}}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\sin\sqrt{\text{x}}}\big)$
$=\text{e}^{\sin\sqrt{\text{x}}}\frac{\text{d}}{\text{dx}}\big(\sin\sqrt{\text{x}}\big)$
[Using chain rule]
$=\text{e}^{\sin\sqrt{\text{x}}}\times\cos\sqrt{\text{x}}\frac{\text{d}}{\text{dx}}\sqrt{\text{x}}$
[Using chain rule]
$=\text{e}^{\sin\sqrt{\text{x}}}\times\cos\sqrt{\text{x}}\times\frac{1}{2\sqrt{\text{x}}}$
$=\frac{1}{2\sqrt{\text{x}}}\times\cos\sqrt{\text{x}}\times\text{e}^{\sin\sqrt{\text{x}}}$
So,
$\frac{\text{d}}{\text{dx}}=\big(\text{e}^{\sin^\sqrt{\text{x}}}\big)=\frac{1}{2\sqrt{\text{x}}}\cos\sqrt{\text{x}}\times\text{e}^{\sin\sqrt{\text{x}}}$

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Maths 3 Marks Question