if f(x)=x + sin x, then f(x) = \na 1 + cos x\nb 1 - cos x\nc cos x\nd sin x - xcos x

Explanation:

Step1: Recall derivative rules

The derivative of a sum of functions $u(x)+v(x)$ is $u'(x)+v'(x)$. Here $u(x)=x$ and $v(x)=\sin x$.

Step2: Differentiate $u(x)$

The derivative of $x$ with respect to $x$ is 1, i.e., $\frac{d}{dx}(x) = 1$.

Step3: Differentiate $v(x)$

The derivative of $\sin x$ with respect to $x$ is $\cos x$, i.e., $\frac{d}{dx}(\sin x)=\cos x$.

Step4: Find $f'(x)$

Since $f(x)=x + \sin x$, then $f'(x)=\frac{d}{dx}(x)+\frac{d}{dx}(\sin x)=1+\cos x$.

Answer:

A. $1+\cos x$