Q: Differentiate the following wrt $x: \cos \left(\log x+e^{x}\right), x>0$

Let $y=\cos \left(\log x+e^{x}\right)$ By using the chain rule, we get $$ \begin{aligned} \frac{d y}{d x} & =-\sin \left[\log x+e^{x}\right] \cdot \frac{d}{d x}\left(\log x+e^{x}\right) \\ & =-\sin \left(\log x+e^{x}\right) \cdot\left[\frac{d}{d x}(\log x)+\frac{d}{d x}\left(e^{x}\right)\right] \\ & =-\sin \left(\log x+e^{x}\right) \cdot\left(\frac{1}{x}+e^{x}\right) \\ & =-\left(\frac{1}{x}+e^{x}\right) \sin \left(\log x+e^{x}\right), x>0 \end{aligned} $$ \section*{EXERCISE 5.5}

Q: Differentiate the following wrt $x: \sin \left(\tan ^{-1} e^{-x}\right)$

Let $y=\sin \left(\tan ^{-1} e^{-x}\right)$ By using the chain rule, we get $$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[\sin \left(\tan ^{-1} e^{-x}\right)\right] \\ & =\cos \left(\tan ^{-1} e^{-x}\right) \cdot \frac{d}{d x}\left(\tan ^{-1} e^{-x}\right) \\ & =\cos \left(\tan ^{-1} e^{-x}\right) \cdot \frac{1}{1+\left(e^{-x}\right)^{2}} \cdot \frac{d}{d x}\left(e^{-x}\right) \\ & =\frac{\cos \left(\tan ^{-1} e^{-x}\right)}{1+e^{-2 x}} \cdot e^{-x} \cdot \frac{d}{d x}(-x) \\ & =\frac{e^{-x} \cos \left(\tan ^{-1} e^{-x}\right)}{1+e^{-2 x}} \times(-1) \\ & =\frac{-e^{-x} \cos \left(\tan ^{-1} e^{-x}\right)}{1+e^{-2 x}} \end{aligned} $$

Q: Differentiate the following wrt $x: \log \left(\cos e^{x}\right)$

Let $y=\log \left(\cos e^{x}\right)$ By using the chain rule, we get $$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[\log \left(\cos e^{x}\right)\right] \\ & =\frac{1}{\cos e^{x}} \cdot \frac{d}{d x}\left(\cos e^{x}\right) \\ & =\frac{1}{\cos e^{x}} \cdot\left(-\sin e^{x}\right) \cdot \frac{d}{d x}\left(e^{x}\right) \\ & =\frac{-\sin e^{x}}{\cos e^{x}} \cdot e^{x} \\ & =-e^{x} \tan e^{x}, e^{x} \neq(2 n+1) \frac{\pi}{2}, n \in \mathbf{N} \end{aligned} $$

Q: Differentiate the following wrt $x: e^{x}+e^{x^{2}}+\ldots+e^{x^{5}}$

$\frac{d}{d x}\left(e^{x}+e^{x^{2}}+\ldots+e^{x^{5}}\right)$ Differentiating wrt $x$, we get $$ \begin{aligned} \frac{d}{d x}\left(e^{x}+e^{x^{2}}+\ldots+e^{x^{5}}\right) & =\frac{d}{d x}\left(e^{x}\right)+\frac{d}{d x}\left(e^{x^{2}}\right)+\frac{d}{d x}\left(e^{x^{3}}\right)+\frac{d}{d x}\left(e^{x^{4}}\right)+\frac{d}{d x}\left(e^{x^{5}}\right) \\ & =e^{x}+\left[e^{x^{2}} \times \frac{d}{d x}\left(x^{2}\right)\right]+\left[e^{x^{3}} \times \frac{d}{d x}\left(x^{3}\right)\right]+\left[e^{x^{4}} \times \frac{d}{d x}\left(x^{4}\right)\right]+\left[e^{x^{5}} \times \frac{d}{d x}\left(x^{5}\right)\right] \\ & =e^{x}+\left(e^{x^{2}} \times 2 x\right)+\left(e^{x^{3}} \times 3 x^{2}\right)+\left(e^{x^{4}} \times 4 x^{3}\right)+\left(e^{x^{5}} \times 5 x^{4}\right) \\ & =e^{x}+2 x e^{x^{2}}+3 x^{2} e^{x^{3}}+4 x^{3} e^{x^{4}}+5 x^{4} e^{x^{5}} \end{aligned} $$

Question 1

Differentiating the following wrt $x: \sqrt{e^{\sqrt{x}}}, x>0$

Accepted Answer

Let $y=\sqrt{e^{\sqrt{x}}}$

Then, $y^{2}=e^{\sqrt{x}}$

Differentiating wrt $x$, we get

$y^{2}=e^{\sqrt{x}}$

$$

\begin{aligned}

& \frac{d}{d x}\left(y^{2}ight)=\frac{d}{d x}\left(e^{\sqrt{x}}ight) \

& \Rightarrow 2 y \frac{d y}{d x}=e^{\sqrt{x}} \frac{d}{d x}(\sqrt{x}) \

& \Rightarrow 2 y \frac{d y}{d x}=e^{\sqrt{x}} \frac{1}{2} \cdot \frac{1}{\sqrt{x}} \

& \Rightarrow \frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 y \sqrt{x}} \

& \Rightarrow \frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 \sqrt{e^{\sqrt{x}}} \sqrt{x}} \

& \Rightarrow \frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 \sqrt{x e^{\sqrt{x}}}}, x>0

\end{aligned}

$$

Question 2

Differentiating the following wrt $x: \log (\log x), x>1$

Accepted Answer

Let $y=\log (\log x)$

By using the chain rule, we get

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}[\log (\log x)] \

& =\frac{1}{\log x} \cdot \frac{d}{d x}(\log x) \

& =\frac{1}{\log x} \cdot \frac{1}{x} \

& =\frac{1}{x \log x}, x>1

\end{aligned}

$$

Question 3

Differentiating the following wrt $x: \frac{\cos x}{\log x}, x>0$

Accepted Answer

Let $y=\frac{\cos x}{\log x}$

By using the quotient rule, we get

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{\frac{d}{d x}(\cos x) \cdot \log x-\cos x \cdot \frac{d}{d x}(\log x)}{(\log x)^{2}} \

& =\frac{-\sin x \log x-\cos x \cdot \frac{1}{x}}{(\log x)^{2}} \

& =-\left[\frac{x \log x \cdot \sin x+\cos x}{x(\log x)^{2}}ight], x>0

\end{aligned}

$$

Question 4

Differentiate the following wrt $x: \cos \left(\log x+e^{x}\right), x>0$

Accepted Answer

Let $y=\cos \left(\log x+e^{x}ight)$

By using the chain rule, we get

$$

\begin{aligned}

\frac{d y}{d x} & =-\sin \left[\log x+e^{x}ight] \cdot \frac{d}{d x}\left(\log x+e^{x}ight) \

& =-\sin \left(\log x+e^{x}ight) \cdot\left[\frac{d}{d x}(\log x)+\frac{d}{d x}\left(e^{x}ight)ight] \

& =-\sin \left(\log x+e^{x}ight) \cdot\left(\frac{1}{x}+e^{x}ight) \

& =-\left(\frac{1}{x}+e^{x}ight) \sin \left(\log x+e^{x}ight), x>0

\end{aligned}

$$

\section*{EXERCISE 5.5}

Question 5

Differentiating the following wrt $x: \frac{e^{x}}{\sin x}$

Accepted Answer

Let $y=\frac{e^{x}}{\sin x}$

By using the quotient rule, we get

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{\sin x \frac{d}{d x}\left(e^{x}ight)-e^{x} \frac{d}{d x}(\sin x)}{\sin ^{2} x} \

& =\frac{\sin x \cdot\left(e^{x}ight)-e^{x} \cdot(\cos x)}{\sin ^{2} x} \

& =\frac{e^{x}(\sin x-\cos x)}{\sin ^{2} x}

\end{aligned}

$$

Question 6

Differentiating the following $e^{\sin ^{-1} x}$

Accepted Answer

Let $y=e^{\sin ^{-1} x}$

By using the quotient rule, we get

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left(e^{\sin ^{-1} x}ight) \

& =e^{\sin ^{-1} x} \cdot \frac{d}{d x}\left(\sin ^{-1} xight) \

& =e^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^{2}}} \

& =\frac{e^{\sin ^{-1} x}}{\sqrt{1-x^{2}}} \

& =\frac{e^{\sin ^{-1} x}}{\sqrt{1-x^{2}}}, x \in(-1,1)

\end{aligned}

$$

Question 7

Differentiating the following wrt $x: e^{x^{3}}$

Accepted Answer

Let $y=e^{x^{3}}$

By using the quotient rule, we get

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left(e^{x^{3}}ight) \

& =e^{x^{3}} \cdot \frac{d}{d x}\left(x^{3}ight) \

& =e^{x^{3}} \cdot 3 x^{2} \

& =3 x^{2} e^{x^{3}}

\end{aligned}

$$

Question 8

Differentiate the following wrt $x: \sin \left(	an ^{-1} e^{-x}ight)$

Accepted Answer

Let $y=\sin \left(	an ^{-1} e^{-x}ight)$

By using the chain rule, we get

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left[\sin \left(	an ^{-1} e^{-x}ight)ight] \

& =\cos \left(	an ^{-1} e^{-x}ight) \cdot \frac{d}{d x}\left(	an ^{-1} e^{-x}ight) \

& =\cos \left(	an ^{-1} e^{-x}ight) \cdot \frac{1}{1+\left(e^{-x}ight)^{2}} \cdot \frac{d}{d x}\left(e^{-x}ight) \

& =\frac{\cos \left(	an ^{-1} e^{-x}ight)}{1+e^{-2 x}} \cdot e^{-x} \cdot \frac{d}{d x}(-x) \

& =\frac{e^{-x} \cos \left(	an ^{-1} e^{-x}ight)}{1+e^{-2 x}} 	imes(-1) \

& =\frac{-e^{-x} \cos \left(	an ^{-1} e^{-x}ight)}{1+e^{-2 x}}

\end{aligned}

$$

Question 9

Differentiate the following wrt $x: \log \left(\cos e^{x}\right)$

Accepted Answer

Let $y=\log \left(\cos e^{x}ight)$

By using the chain rule, we get

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left[\log \left(\cos e^{x}ight)ight] \

& =\frac{1}{\cos e^{x}} \cdot \frac{d}{d x}\left(\cos e^{x}ight) \

& =\frac{1}{\cos e^{x}} \cdot\left(-\sin e^{x}ight) \cdot \frac{d}{d x}\left(e^{x}ight) \

& =\frac{-\sin e^{x}}{\cos e^{x}} \cdot e^{x} \

& =-e^{x} 	an e^{x}, e^{x} 
eq(2 n+1) \frac{\pi}{2}, n \in \mathbf{N}

\end{aligned}

$$

Question 10

Differentiate the following wrt $x: e^{x}+e^{x^{2}}+\ldots+e^{x^{5}}$

Accepted Answer

$\frac{d}{d x}\left(e^{x}+e^{x^{2}}+\ldots+e^{x^{5}}ight)$

Differentiating wrt $x$, we get

$$

\begin{aligned}

\frac{d}{d x}\left(e^{x}+e^{x^{2}}+\ldots+e^{x^{5}}ight) & =\frac{d}{d x}\left(e^{x}ight)+\frac{d}{d x}\left(e^{x^{2}}ight)+\frac{d}{d x}\left(e^{x^{3}}ight)+\frac{d}{d x}\left(e^{x^{4}}ight)+\frac{d}{d x}\left(e^{x^{5}}ight) \

& =e^{x}+\left[e^{x^{2}} 	imes \frac{d}{d x}\left(x^{2}ight)ight]+\left[e^{x^{3}} 	imes \frac{d}{d x}\left(x^{3}ight)ight]+\left[e^{x^{4}} 	imes \frac{d}{d x}\left(x^{4}ight)ight]+\left[e^{x^{5}} 	imes \frac{d}{d x}\left(x^{5}ight)ight] \

& =e^{x}+\left(e^{x^{2}} 	imes 2 xight)+\left(e^{x^{3}} 	imes 3 x^{2}ight)+\left(e^{x^{4}} 	imes 4 x^{3}ight)+\left(e^{x^{5}} 	imes 5 x^{4}ight) \

& =e^{x}+2 x e^{x^{2}}+3 x^{2} e^{x^{3}}+4 x^{3} e^{x^{4}}+5 x^{4} e^{x^{5}}

\end{aligned}

$$