3 Marks

Question 513 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 523 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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Question 533 Marks

Differentiate the following functions with respect to x:
$\tan5\text{x}^\circ$

Answer

Let, $\text{y}=\tan5\text{x}^\circ$
$\Rightarrow\ \text{y}=\tan\Big(5\text{x}\times\frac{\pi}{180}\Big)$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\tan\Big(5\text{x}\times\frac{\pi}{180}\Big)$
$=\sec^2\Big(5\text{x}\times\frac{\pi}{180}\Big)\frac{\text{d}}{\text{dx}}\Big(5\text{x}\times\frac{\pi}{108}\Big) $
[Using chain rule]
$=\Big(\frac{5\text{x}}{180}\Big)\sec^2\Big(5\text{x}\times\frac{\pi}{180}\Big)$
$=\frac{5\pi}{180}\sec^2(5\text{x}^\circ)$
Hence, $\frac{\text{d}}{\text{dx}}(\tan5\text{x}^\circ)=\frac{5\pi}{180}\sec^2(5\text{x}^\circ)$

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Question 543 Marks

Differentiate the following functions from first principles:
e^3x.

Answer

Let f(x) = e^3x
⇒ f(x + h) = e^{3(x + h)}
$\frac{\text{d}}{\text{dx}}(\text{f(x)})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{3(\text{x}+\text{h})}-\text{e}^{3\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{3\text{x}}\text{e}^{3\text{h}}-\text{e}^{3\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{e}^{3\text{x}}\left\{\frac{(\text{e}^{3\text{h}}-1)}{3\text{h}}\right\}\times3$
$=3\text{e}^{3\text{x}}\Big[\text{Since, }\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}=1\Big]$
Hence,
$\frac{\text{d}}{\text{dx}}(\text{e}^{3\text{x}})=3\text{e}^{3\text{x}}$

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Question 553 Marks

Differentiate the following functions with respect to x:
$\sin^2(2\text{x}+1)$

Answer

Cobnsider $\text{y}=\sin^2(2\text{x}+1)$
Differentiate it with respect to x,
$\frac{\text{d}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\sin^2(2\text{x}+1)\big]$
$=2\sin(2\text{x}+1)\frac{\text{d}}{\text{dx}}\sin(2\text{x}+1)$
[Using chain rule]
$=2\sin(2\text{x}+1)\cos(2\text{x}+1)\frac{\text{d}}{\text{dx}}\sin(2\text{x}+1)$
[Using chain rule]
$=4\sin(2\text{x}+1)\cos(2\text{x}+1)$
$=2\sin(2\text{x}+1)$
$\Big[\text{Since}, \sin^2\text{A}=2\sin\text{A}\cos\text{A}\Big]$
$2\sin(4\text{x}+2)$
Hence, the solution is $\frac{\text{d}}{\text{dx}}\big(\sin^2(2\text{x}+1)\big)=2\sin(4\text{x}+2)$

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Question 563 Marks

Differentiate the following functions from first principles:
e^-x.

Answer

Consider f(x) = e^-x
⇒ f(x + h) = e^-(x+h)
$\frac{\text{d}}{\text{dx}}(\text{f}(\text{x}))=\frac{\lim}{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\frac{\lim}{\text{h}\rightarrow0}\frac{\text{e}^{-(\text{x}+\text{h})}\text{e}^{-\text{x}}}{\text{h}}$
$=\frac{\lim}{\text{h}\rightarrow0}\frac{\text{e}^{-\text{x}}\times\text{e}^{-\text{h}}-\text{e}^{-\text{x}}}{\text{h}}$
$=\frac{\lim}{\text{h}\rightarrow0}\text{e}^{-\text{x}}\left\{\Big(\frac{\text{e}^{-\text{h}}-1}{-\text{h}}\Big)\right\}\times(-1)$
$\Big[\text{Since, }\frac{\lim}{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{\text{h}}=1\Big]$
$=-\text{e}^{-\text{x}}$
So,
$\frac{\text{d}}{\text{dx}}(\text{e}^{-\text{x}})=-\text{e}^{-\text{x}}$

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Question 573 Marks

If $\text{y}=\text{x}\sin\text{y},$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\sin^2\text{y}}{(1-\text{x}\cos\text{y})}$

Answer

Here,
$\text{y}=\text{x}\sin\text{y}$
Differentiate with respect to x,
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\sin\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\sin\text{y})+\sin\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
[Using product rule]
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\cos\frac{\text{dy}}{\text{dx}}+\sin\text{y}(1)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(1-\text{x}\cos\text{y})=\sin\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin\text{y}}{1-\text{x}\cos\text{y}}$

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Question 583 Marks

Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$\tan^{-1}\big(\text{x}^2+\text{y}^2\big)=\text{a}$

Answer

We have, $\tan^{-1}\big(\text{x}^2+\text{y}^2\big)=\text{a}$

Differentiating with respect to x, we get,

$\frac{\text{d}}{\text{dx}}\big[\tan^{-1}\big(\text{x}^2+\text{y}^2\big)\big]=\frac{\text{d}}{\text{dx}}(\text{a})$

$\Rightarrow\frac{1}{1+(\text{x}^2+\text{y}^2)^2}\times\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\text{y}^2\big)=0$

$\Rightarrow\Big[\frac{1}{1+(\text{x}^2+\text{y}^2)^2}\Big]\Big(2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big)=0$

$\Rightarrow2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}}{\text{y}}$

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Maths 3 Marks