Exercise 5.4 (Revised) - Chapter 5 - Continuity & Differentiability - Ncert Solutions class 12 - Maths
Updated On 26-08-2025 By Lithanya
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Chapter 5: Continuity & Differentiability NCERT Solutions Class 12 Maths
Ex 5.4 Question 1 :
Differentiate the following w.r.t. $x$.
$
\frac{e^x}{\sin x}
$
Answer :
Let $y=\frac{e^x}{\sin x}$
By using the quotient rule, we obtain
$
\begin{aligned}
\frac{d y}{d x} & =\frac{\sin x \frac{d}{d x}\left(e^x\right)-e^x \frac{d}{d x}(\sin x)}{\sin ^2 x} \\
& =\frac{\sin x \cdot\left(e^x\right)-e^x \cdot(\cos x)}{\sin ^2 x} \\
& =\frac{e^x(\sin x-\cos x)}{\sin ^2 x}, x \neq n \pi, n \in \mathbf{Z}
\end{aligned}
$
Ex 5.4 Question 2:
Differentiate the following w.r.t. $x$.
$
e^{\sin ^{-1} x}
$
Answer :
Let $y=e^{\sin ^{-1} x}$
By using the chain rule, we obtain
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}\left(e^{\sin ^{-1} x}\right) \\
& \Rightarrow \frac{d y}{d x}=e^{\sin ^{-1} x} \cdot \frac{d}{d x}\left(\sin ^{-1} x\right) \\
& =e^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}} \\
& =\frac{e^{\sin ^{-1} x}}{\sqrt{1-x^2}} \\
& \therefore \frac{d y}{d x}=\frac{e^{\sin ^{-1} x}}{\sqrt{1-x^2}}, x \in(-1,1) \\
&
\end{aligned}
$
Ex 5.4 Question 3 :
Differentiate the following w.r.t. $x$.
$e^{x^3}$
Answer :
Let $y=e^{x^3}$
By using the chain rule, we obtain
$
\frac{d y}{d x}=\frac{d}{d x}\left(e^{x^3}\right)=e^{x^3} \cdot \frac{d}{d x}\left(x^3\right)=e^{x^3} \cdot 3 x^2=3 x^2 e^{x^3}
$
Ex 5.4 Question 4:
Differentiate the following w.r.t. $x$.
$
\sin \left(\tan ^{-1} e^{-x}\right)
$
Answer:
Let $y=\sin \left(\tan ^{-1} e^{-x}\right)$
By using the chain rule, we obtain
$
\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}
$
Ex 5.4 Question 5 :
Differentiate the following w.r.t. $x$.
$
\log \left(\cos e^x\right)
$
Answer :
Let $y=\log \left(\cos e^x\right)$
By using the chain rule, we obtain
$
\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}
$
Ex 5.4 Question 6:
Differentiate the following w.r.t. $x$.
$
e^x+e^{x^2}+\ldots+e^{x^3}
$
Answer :
$
\frac{d}{d x}\left(e^x+e^{x^2}+\ldots+e^{x^2}\right)
$
$\begin{aligned}
& =\frac{d}{d x}\left(e^x\right)+\frac{d}{d x}\left(e^{x^2}\right)+\frac{d}{d x}\left(e^{x^3}\right)+\frac{d}{d x}\left(e^{x^4}\right)+\frac{d}{d x}\left(e^{x^5}\right) \\
& =e^x+\left[e^{x^2} \times \frac{d}{d x}\left(x^2\right)\right]+\left[e^{x^3} \cdot \frac{d}{d x}\left(x^3\right)\right]+\left[e^{x^4} \cdot \frac{d}{d x}\left(x^4\right)\right]+\left[e^{x^3} \cdot \frac{d}{d x}\left(x^5\right)\right] \\
& =e^x+\left(e^{x^2} \times 2 x\right)+\left(e^{x^3} \times 3 x^2\right)+\left(e^{x^4} \times 4 x^3\right)+\left(e^{x^3} \times 5 x^4\right) \\
& =e^x+2 x e^{x^2}+3 x^2 e^{x^3}+4 x^3 e^{x^4}+5 x^4 e^{x^3}
\end{aligned}$
Ex 5.4 Question 7
Differentiate the following w.r.t. $x$.
$
\sqrt{e^{\sqrt{x}}}, x>0
$
Answer :
Let $y=\sqrt{e^{\sqrt{x}}}$
Then, $y^2=e^{\sqrt{x}}$
By differentiating this relationship with respect to $x$, we obtain
$
\begin{aligned}
& y^2=e^{\sqrt{x}} \\
& \Rightarrow 2 y \frac{d y}{d x}=e^{\sqrt{x}} \frac{d}{d x}(\sqrt{x}) \quad \text { [By applying the chain rule] } \\
& \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}
$
Ex 5.4 Question 8:
Differentiate the following w.r.t. $x$.
$
\log (\log x), x>1
$
Answer :
Let $y=\log (\log x)$
By using the chain rule, we obtain
$
\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}
$
Ex 5.4 Question 9 :
Differentiate the following w.r.t. $x$.
$
\frac{\cos x}{\log x}, x>0
$
Answer :
Let $y=\frac{\cos x}{\log x}$
By using the quotient rule, we obtain
$
\begin{aligned}
\frac{d y}{d x} & =\frac{\frac{d}{d x}(\cos x) \times \log x-\cos x \times \frac{d}{d x}(\log x)}{(\log x)^2} \\
& =\frac{-\sin x \log x-\cos x \times \frac{1}{x}}{(\log x)^2} \\
& =\frac{-[x \log x \cdot \sin x+\cos x]}{x(\log x)^2}, x>0
\end{aligned}
$
Ex 5.4 Question 10:
Differentiate the following w.r.t. $x$.
$
\cos \left(\log x+e^x\right), x>0
$
Answer :
Let $y=\cos \left(\log x+e^x\right)$
By using the chain rule, we obtain
$
\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}
$
