Question

$\frac{d}{dx}(\cos x\tan x)=$
a $sec x+sin x\tan x$
b $cos x$
c $-sin xsec^{2}x$
d $sin x$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Let $u = \cos x$ and $v=\tan x$.

Step2: Find $u^\prime$ and $v^\prime$

The derivative of $u=\cos x$ is $u^\prime=-\sin x$, and the derivative of $v = \tan x=\frac{\sin x}{\cos x}$ is $v^\prime=\sec^{2}x$.

Step3: Calculate the derivative

$\frac{d}{dx}(\cos x\tan x)=(-\sin x)\tan x+\cos x\sec^{2}x$. Since $\sec x=\frac{1}{\cos x}$, then $\cos x\sec^{2}x=\sec x$. So $\frac{d}{dx}(\cos x\tan x)=\sec x-\sin x\tan x$.

Answer:

D. $\sin x$