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Class 12 Maths Continuity and Differentiability NCERT Solutions Exercise 5.4




In this page we have Class 12 Maths Continuity and Differentiability NCERT Solutions Exercise 5.4 . Hope you like them and do not forget to like , social share and comment at the end of the page.

Differentiate the following w.r.t. x


 
Question 1
$\frac{e^{x}}{\sin x}$
Solution
Let y=$\frac{e^{x}}{\sin x}$
Using quotient rule, we have
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\sin x.\frac{\mathrm{d} }{\mathrm{d} x}(e^{x})-e^{x}.\frac{\mathrm{d} }{\mathrm{d} x}(\sin x)}{\sin^{2} x}$
$=\frac{\sin x.(e^{x})-e^{x}.(\cos x)}{\sin^{2} x}$
$=\frac{e^{x}(\sin x-\cos x)}{\sin^{2} x},x\neq n\pi ,n\in \mathbb{Z}$  

Question 2
$e^{\sin ^{-1}x}$
Solution

Let y=$y=e^{\sin ^{-1}x}$
Using chain rule, we have
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}(e^{\sin ^{-1}x})$
$ \frac{\mathrm{d} y}{\mathrm{d} x}=e^{\sin^{-1}x}.\frac{\mathrm{d} }{\mathrm{d} x}(\sin^{-1}x)$
$=e^{\sin^{-1}x}.\frac{1}{\sqrt{1-x^{2}}}$
$=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}}$
$\frac{\mathrm{d} y}{\mathrm{d}x}=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}},x\in (-1,1)$
 
 
Question 3
$e^{x^{3}}$
Solution
Let y=$e^{x^{3}}$
Using chain rule, we have
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}(e^{x^{3}})=e^{x^{3}}.3x^{2}=3x^{2}.e^{x^{3}}$  
 
Question 4
$\sin (\tan^{-1} e^{-x})$
Solution
Let y=$\sin (\tan^{-1} e^{-x})$
Using chain rule, we have
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}[\sin (\tan^{-1} e^{-x})]$
$=\cos (\tan^{-1} e^{-x}).\frac{\mathrm{d} }{\mathrm{d} x}(\tan^{-1} e^{-x})$
$=\cos (\tan^{-1} e^{-x}).\frac{1}{1+(e^{-x})^{2}}(\tan^{-1} e^{-x}).\frac{\mathrm{d} }{\mathrm{d} x}(e^-{x})$
$=\frac{\cos (\tan^{-1} e^{-x})}{1+e^{-2x}}.e^{-x}.\frac{\mathrm{d} }{\mathrm{d} x}(-x)$
$=-\frac{\cos (\tan^{-1} e^{-x})}{1+e^{-2x}}.e^{-x}$

Question 5
$log(\cos e^{x})$
Solution
Let y=$log(\cos e^{x})$
Using chain rule, we have
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}[log(\cos e^{x})]$
$=\frac {1}{\cos e^{x}} .\frac{\mathrm{d} }{\mathrm{d} x}(\cos e^{x})$
$=\frac {1}{\cos e^{x}} .(-\sin e^{x}) \frac{\mathrm{d} }{\mathrm{d} x}(e^{x})$
$=-\frac{\sin e^{x}}{\cos e^{x}}.(e^{x})$
$=-\tan e^x.(e^{x})$



Question 6
$e^{x} +e^{x^2}+e^{x^3}+e^{x^4}+e^{x^5}$
Solution
Let y=$e^{x} +e^{x^2}+e^{x^3}+e^{x^4}+e^{x^5}$
Using chain rule, we have
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}[e^{x} +e^{x^2}+e^{x^3}+e^{x^4}+e^{x^5}]$
$=\frac{\mathrm{d} }{\mathrm{d} x} e^{x} + \frac{\mathrm{d} }{\mathrm{d} x}e^{x^2}+ \frac{\mathrm{d} }{\mathrm{d} x} e^{x^3}+ \frac{\mathrm{d} }{\mathrm{d} x} e^{x^4}+\frac{\mathrm{d} }{\mathrm{d} x}e^{x^5}$
$=e^{x} +2xe^{x^2}+3x^2 e^{x^3}+4x^3 e^{x^4}+5x^4 e^{x^5}$

Question 7
$\sqrt{e^ {\sqrt x}}$
Solution
Let $y=\sqrt{e^ {\sqrt x}}$
Then $y^{2}=e^ {\sqrt x}$ Using chain rule, we have
$2y\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}[e^ {\sqrt x}]$
$2y\frac{\mathrm{d} y}{\mathrm{d} x}=e^ {\sqrt x}\frac{\mathrm{d} }{\mathrm{d} x} \sqrt x$
$2y\frac{\mathrm{d} y}{\mathrm{d} x}=e^ {\sqrt x} \frac {1}{2\sqrt x}$
$\frac{\mathrm{d} y}{\mathrm{d} x}=e^ {\sqrt x} \frac {1}{4y\sqrt x}$
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac {e^ {\sqrt x}}{4\sqrt{xe^ {\sqrt x}}}$

Question 8
$log(log x), x > 1$
Solution
Let $y=log(log x)$
Using chain rule, we have
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{log x}\frac{\mathrm{d} }{\mathrm{d} x} log x$
$=\frac{1}{log x}\frac{1}{x}$
$=\frac{1}{xlog x}$


 


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