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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 29 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 29 Maths Textbook Solution.

Answers (1)

Answer:\frac{dy}{dx}=\frac{a}{a^{2}+x^{2}}

Hint:

\begin{aligned} &\frac{d}{d x}(\text { constan } t)=0 ; \\ &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \end{aligned}

Given:

\tan ^{-1}\left ( \frac{x-a}{x+a} \right )

Solution:

Let,

\begin{aligned} &y=\tan ^{-1}\left(\frac{x-a}{x+a}\right) \\ &y=\tan ^{-1}\left(\frac{\frac{x-a}{x}}{\frac{x+a}{x}}\right) \end{aligned}

\begin{aligned} &y=\tan ^{-1}\left(\frac{\frac{x}{x}-\frac{a}{x}}{\frac{x}{x}+\frac{a}{x}}\right) \\ &y=\tan ^{-1}\left(\frac{1-\frac{a}{x}}{1+1 \times \frac{a}{x}}\right) \\ &y=\tan ^{-1}(1)+\tan ^{-1}\left(\frac{a}{x}\right) \end{aligned}

Using,

\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)

Differentiating its with respect to x using chain tule.

\begin{aligned} &\frac{d y}{d x}=0-\frac{1}{1+\left(\frac{a}{x}\right)^{2}} \frac{d}{d x}\left(\frac{a}{x}\right) \\ &\frac{d y}{d x}=\frac{-1}{1+\frac{a^{2}}{x^{2}}}\left(-\frac{a}{x^{2}}\right) \\ &\frac{d y}{d x}=\frac{-1}{\frac{x^{2}+a^{2}}{x^{2}}} \times\left(\frac{-a}{x^{2}}\right) \\ &\frac{d y}{d x}=\frac{-x^{2}}{x^{2}+a^{2}} \times\left(\frac{-a}{x^{2}}\right) \end{aligned}                                                                                                            Using   \frac{d}{dx}\left ( x^{n} \right )=nx^{n-1}

\frac{d}{dx}=\frac{a}{x^{2}+a^{2}}                                                                                                                    \frac{d}{dx}\left ( constant \right )=0

 

 

 

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