Question

If $y={\tan }^{-1}\left(\frac{4x}{1+5x^2}\right)+{\cot }^{-1}\left(\frac{3-2x}{2+3x}\right)$ then $\frac{dy}{dx}$ is

• A. $\frac{-5}{1+25x^2}$
• B. $\frac{5}{1+25x^2}$
• C. $\frac{5}{1-25x^2}$
• D. $0$

Answer 1

Answer: (B)

$\frac{5}{1+25x^2}$

As we know that

${\tan }^{-1}(a+b)=\frac{{\tan }^{-1}a+{\tan }^{-1}b}{1-{\tan }^{-1}a{\tan }^{-1}b}$

And similarly,

$\Rightarrow {\tan }^{-1}(a-b)=\frac{{\tan }^{-1}a-{\tan }^{-1}b}{1+{\tan }^{-1}a{\tan }^{-1}b}$

$\Rightarrow {\tan }^{-1}\left(\frac{4x}{1+5x^2}\right)={\tan }^{-1}\left(\frac{5x-x}{1+5x.x}\right)={\tan }^{-1}5x-{\tan }^{-1}x$

$\Rightarrow {\cot }^{-1}x={\tan }^{-1}\left(\frac{1}{x}\right)$

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

$\Rightarrow \therefore y={\tan }^{-1}5x+{\tan }^{-1}\left(\frac{2}{3}\right)$

$\Rightarrow \frac{dy}{dx}=\frac{5}{1+25x^2}$