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 $y^\prime=u^\prime v + uv^\prime$. 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. $\cos x$