Class 12 ncert solution math exercise 5.4

The  Class 12 NCERT solution math exercise 5.4 prepared by expert Mathematics teacher at as per CBSE  guidelines. See our Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4 Questions with Solutions to help you to revise complete Syllabus and Score More marks in your Board and School exams.

    EXERCISE 5.4 ( Differentiation )

Differentiating the following w.r.t.(Class 12 ncert solution math exercise 5.4)

Question 1: \frac{e^x}{\sin x}.

Solution:Let y=\frac{e^x}{\sin x}

By using the quotient rule, we get

\frac{dy}{dx} =\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}

Question 2: e^{\sin ^{-1} x}

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

By using the quotient rule, we get

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

=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}}

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

Question 3: e^{x^3}

Solution:Let y=e^{x^3}

By using the quotient rule, we get

\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}

Question 4:  \sin \left(\tan ^{-1} e^{-x}\right)

Solution:Let y=\sin \left(\tan ^{-1} e^{-x}\right)

By using the chain rule, we get

\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}}

Question 5: \log \left(\cos e^x\right)

Solution:Let y=\log \left(\cos e^x\right)

By using the chain rule, we get

\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}

Question 6: e^x+e^{x^2}+\ldots+e^{x^5}

Solution:\frac{d}{d x}\left(e^x+e^{x^2}+\ldots+e^{x^5}\right)

Differentiating wrt x, we get

\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}{dx}\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}

Question 7: \sqrt{e^{\sqrt{x}}}, x>0

Solution: Let y=\sqrt{e^{\sqrt{x}}}

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

Differentiating wrt x, we get


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

\Rightarrow 2 y \frac{dy}{dx}=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

Question 8: \log (\log x), x>1

Solution: Let y=\log (\log x)

By using the chain rule, we get

\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

Question 9:  \frac{\cos x}{\log x}, x>0

Solution: Let y=\frac{\cos x}{\log x}

By using the quotient rule, we get

\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}\right], x>0

Question 10: \cos \left(\log x+e^x\right), x>0

Solution: Let y=\cos \left(\log x+e^x\right)

By using the chain rule, we get

\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

NCERT solution chapter 5 continuity and differentiability

1. Ncert solution  Exercise 5.1 continuity and differentiability

2. Ncert solution  Exercise 5.2 continuity and differentiability

3. Ncert solution  Exercise 5.3 continuity and differentiability

4. Ncert solution  Exercise 5.5 continuity and differentiability

5. Ncert solution  Exercise 5.6 continuity and differentiability

6. Ncert solution  Exercise 5.7 continuity and differentiability

7. Ncert solution  Chapter 5 Miscellaneous continuity and differentiability