Question

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

• A. $\frac{6}{1+4x^2}$
• B. $\frac{3}{1+4x^2}$
• C. $\frac{5}{1+{25x}^2}$
• D. $\frac{5}{(1+25x^2)}-\frac{1}{(1+x^2)}-\frac{1.5}{(1+2.25x^2)}$

Answer 1

The correct option is D $\frac{5}{(1+25x^2)}-\frac{1}{(1+x^2)}-\frac{1.5}{(1+2.25x^2)}$

Use properties of inverse trigonometric functions-

y= ${\tan }^{-1}5x+{\tan }^{-1}(-x)$ +${\tan }^{-1}1+{\tan }^{-1}(-1.5x)$

$\Rightarrow$ y= ${\tan }^{-1}5x-{\tan }^{-1}x$ +${\tan }^{-1}1-{\tan }^{-1}1.5x$

Differentiate with respect to x,

$\therefore$ $\frac{d\left(ta{n}^{-1}x\right)}{dx}$ = $\frac{1}{(1+x^2)}$

So,

$\therefore$ $\frac{dy}{dx}=\frac{5}{(1+25x^2)}-\frac{1}{(1+x^2)}+0-\frac{1.5}{(1+2.25x^2)}$

$\therefore$ $\frac{dy}{dx}=\frac{5}{(1+25x^2)}-\frac{1}{(1+x^2)}-\frac{1.5}{(1+2.25x^2)}$