$\\frac{d}{dx}(\\cos x\\tan x)=$\n\na $sec x+sin x\tan x$\n\nb $cos x$\n\nc $-sin xsec^{2}x$\n\nd $sin x$

Explanation:

Step1: Apply product - rule

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

Step2: Find derivatives of $u$ and $v$

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

Step3: Substitute into product - rule

$\frac{d}{dx}(\cos x\tan x)=(-\sin x)\tan x+\cos x\sec^{2}x$. 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:

There seems to be an error in the provided options as the correct derivative $\frac{d}{dx}(\cos x\tan x)=\sec x - \sin x\tan x$ is not among them. If we simplify further: [ \begin{align*} \frac{d}{dx}(\cos x\tan x)&=\frac{d}{dx}(\cos x\cdot\frac{\sin x}{\cos x})\ &=\frac{d}{dx}(\sin x)\ &=\cos x \end{align*} ] So the answer is B. $\cos x$