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  • Provide solution RD Sharma maths class 12 chapter 10 differentiability exercise 10.1 question 7 maths textbook solution

Provide solution RD Sharma maths class 12 chapter 10 differentiability exercise 10.1 question 7 maths textbook solution

Answers (1)

Ex_10.1_Q_7

Answer:

\frac{e^{\sqrt{\cot x}} \times \operatorname{cosec}^{2} x}{2 \sqrt{\cot x}}

Hint:

Use first principle formula to find the differentiation

Given:

e^{\sqrt{\cot x}}

Solution:

Let,

\begin{aligned} &f(x)=e^{\sqrt{\cot x}} \\ &f(x+h)=e^{\sqrt{\cot (x+h)}} \end{aligned}

Now, we will use the formula of first principle to find the differentiation

\begin{aligned} &\frac{d}{d x}(f(x))=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &\frac{d}{d x}\left(e^{\sqrt{\cot x}}\right)=\lim _{h \rightarrow 0} \frac{e^{\sqrt{\cot (x+h)}}-e^{\sqrt{\cot x}}}{h} \end{aligned}

Take e^{\sqrt{\cot x}} common from numerator

=\lim _{h \rightarrow 0} \frac{e^{\sqrt{\cot x}}\left(\frac{e^{\sqrt{\cot (x+h)}}}{e^{\sqrt{\cot x}}}-1\right)}{h}

=\lim _{h \rightarrow 0} \frac{e^{\sqrt{\cot x}}\left(e^{\sqrt{\cot (x+h)}-\sqrt{\cot x}}-1\right)}{h}                                   \left[\because \frac{a^{m}}{a^{n}}=a^{m-n}\right]

\because e^{\sqrt{\cot x}} Is a function of variable x and limit is applied on variable ‘h’

So,

=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0} \frac{e^{(\sqrt{\cot (x+h)}-\sqrt{\cot x})}-1}{h}

Multiply and divide by  (\sqrt{\cot (x+h)}-\sqrt{\cot x})                     

=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0} \frac{\left[e^{\sqrt{\cot (x+h)}-\sqrt{\cot x}}-1\right]}{(\sqrt{\cot (x+h)}-\sqrt{\cot x})} \times\left[\frac{\sqrt{\cot (x+h)}-\sqrt{\cot x}}{h}\right]

=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0}(1) \times\left[\frac{\sqrt{\cot (x+h)}-\sqrt{\cot x}}{h}\right]                                           \left[\because \lim _{h \rightarrow 0} \frac{e^{x}-1}{x}=1\right]

Rationalising the numerator

=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0}\left[\frac{\sqrt{\cot (x+h)}-\sqrt{\cot x}}{h} \times \frac{\sqrt{\cot (x+h)}+\sqrt{\cot x}}{\sqrt{\cot (x+h)}+\sqrt{\cot x}}\right]

=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0} \frac{\cot (x+h)-\cot x}{h(\sqrt{\cot (x+h)}+\sqrt{\cot x})}                                          \left[\because(a+b)(a-b)=a^{2}-b^{2}\right]\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]

=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0} \frac{\frac{\cot (x+h) \times \cot x+1}{\cot (x+h-x)}}{h(\sqrt{\cot (x+h)}+\sqrt{\cot x})}                            \left[\cot (A+B)=\frac{\cot A \cot B+1}{\cot A-\cot B}\right]

=\left(e^{\sqrt{\cot x}}\right) \times \lim _{h \rightarrow 0} \frac{\cot (x+h) \cot x+1}{h(\operatorname{coth})(\sqrt{\cot (x+h)}+\sqrt{\cot x})}

=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0} \frac{\cot (x+h) \cot x+1}{\left(\frac{h}{\tanh }\right)(\sqrt{\cot (x+h)}+\sqrt{\cot x})}                         \left[\because \tan \theta=\frac{1}{\cot \theta}\right]

 

=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0} \frac{\tanh }{h} \times\left[\frac{\cot (x+h) \cot x+1}{\sqrt{\cot (x+h)}+\sqrt{\cot x}}\right]

=e^{\sqrt{\cot x}} \times 1 \times \lim _{h \rightarrow 0} \frac{\cot (x+h) \cot x+1}{\sqrt{\cot (x+h)}+\sqrt{\cot x}}                              \left[\because \lim _{x \rightarrow 0} \frac{\tan x}{x}=1\right]

=e^{\sqrt{\cot x}} \times \frac{\cot ^{2} x+1}{2 \sqrt{\cot x}}

=e^{\sqrt{\cot x}} \times \frac{\cos e c^{2} x}{2 \sqrt{\cot x}}                                                    \left[\because 1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta\right]

\therefore \frac{d}{d x}\left(e^{\sqrt{\cot x}}\right)=\frac{e^{\sqrt{\cot x}} \times \operatorname{cosec}^{2} x}{2 \sqrt{\cot x}}

Hence, the differentiation of e^{\sqrt{\cot x}} is \frac{e^{\sqrt{\cot x}} \times \operatorname{cosec}^{2} x}{2 \sqrt{\cot x}}

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