Question

Find the derivative of $(x+\cos x)(x-\tan x)$.

Answer 1

Let $f\left(x\right)=(x+\cos x)(x-\tan x)$
Differentiating with respect to $x$
$\Rightarrow f^{\prime }\left(x\right)=\frac{d}{dx}\left(\right(x+\cos x\left)\right(x-\tan x\left)\right)$
$\Rightarrow f^{\prime }\left(x\right)=(x-\tan x)\frac{d}{dx}(x+\cos x)+(x+\cos x)\frac{d}{dx}(x-\tan x)$
$\Rightarrow f^{\prime }\left(x\right)=(x-\tan x)\left(\frac{d}{dx}\right(x)+\frac{d}{dx}\cos x)+(x+\cos x)\left(\frac{d}{dx}\right(x)-\frac{d}{dx}\tan x)$
$\Rightarrow f^{\prime }\left(x\right)=(x-\tan x)(1-\sin x)+(x+\cos x)(1-{\sec }^{2}x)$
$\Rightarrow f^{\prime }\left(x\right)=(1-\sin x)(x-\tan x)-({\sec }^{2}x-1)(x+\cos x)$
$\therefore f^{\prime }\left(x\right)=(1-\sin x)(x-\tan x)-{\tan }^{2}x(x+\cos x)$