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  • Explain Solution R.D.Sharma Class 12 Chapter 10 Differentiation Exercise 10.3 Question 30 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 10 Differentiation  Exercise 10.3 Question 30 Maths Textbook Solution.

Answers (1)

Answer: \frac{dy}{dx}=\frac{3}{1+9x^{2}}-\frac{2}{1+4x^{2}}

Hint:

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

Given:

\tan ^{-1}\left(\frac{x}{1+6 x^{2}}\right)

Solution:

Let, y=\tan ^{-1}\left(\frac{x}{1+6 x^{2}}\right)

y=\tan ^{-1}\left(\frac{3 x-2 x}{1+(3 x)(2 x)}\right)

Since,3x-2x=x

\begin{aligned} &y=\tan ^{-1} 3 x-\tan ^{-1} 2 x \\ &{\left[\text { Since }+\tan x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]} \end{aligned}

Differentiating it with respect tox using chain rule,

\begin{aligned} &\frac{d y}{d x}=\frac{1}{1+\left(3 x^{2}\right)} \frac{d}{d x}(3 x)-\frac{1}{1+(2 x)^{2}} \frac{d}{d x}(2 x) \\ &\left\{\text { Since } \Rightarrow \frac{\mathrm{d} }{\mathrm{d} x}\tan ^{-1}(x)=\frac{1}{1+x^{2}}\right\} \\ &\frac{d y}{d x}=\frac{1}{1+9 x^{2}}(3)-\frac{1}{1+4 x^{2}}(2) \\ &\frac{d y}{d x}=\frac{3}{1+9 x^{2}}-\frac{2}{1+4 x^{2}} \end{aligned}

 

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