Question

If $y={\tan }^{-1}\left(\frac{4\sin 2x}{\cos 2x-6{\sin }^{2}x}\right)$ then $\frac{dy}{dx}$ at $x=0$ is

• A. $10$
• B. $12$
• C. $6$
• D. $8$

Answer 1

The correct option is D $8$

$\frac{d}{dx}\left(ta{n}^{-1}\right(\frac{4sin2x}{cos2x-6si{n}^{2}x}\left)\right)|x=0$

$\Rightarrow \frac{d}{dx}\left(ta{n}^{-1}a\right)\frac{d}{dx}\left(\frac{4sin2x}{cos2x-6si{n}^{2}x}\right)$ $\{Letu=\frac{4sin2x}{cos2x-6si{n}^{2}x}\}$

$\Rightarrow \left(\frac{1}{1+u^2}\right)\frac{4\left[\right(cos2x-6si{n}^{2}x\left)\right(2cos2x)-2x(-2sin2x-123sinxcosx\left)\right]}{(cos2x-6si{n}^{2}x{)}^2}$

$\Rightarrow \frac{1}{1+(\frac{4sin\left(2x\right)}{cos2x-6si{n}^{2}x}{)}^2}\left[\frac{4\left(2cos\right(2x\left)\right(cos2x-6si{n}^{2}x+8si{n}^{2}2x\left)\right)}{(cos2x-6si{n}^2(x){)}^2}\right]$

at $x=0$

$\Rightarrow \left(\frac{1}{1+(\frac{4.0}{1-6\left(0\right)}{)}^2}\right)\left[\frac{4\left(2.\right(1\left)\right(1-6\left(0\right)+8\left(0\right))}{(1-6(0){)}^2}\right]$

$=\left(\frac{1}{1+0}\right)\left[\frac{4\left(2\right(1-0)+0)}{1}\right]$

$1\left(\mathrm{4.2.1}\right)=8$ Option D