Question

$$\frac{d}{dx}(x - 8)sin(x) = square$$

Explanation:

Step1: Use the difference - rule of differentiation

The derivative of a difference of functions $u(x)-v(x)$ is $u'(x)-v'(x)$. Here $u(x) = 8\sin(x)$ and $v(x)=x$.

Step2: Differentiate $u(x) = 8\sin(x)$

The derivative of $\sin(x)$ is $\cos(x)$. By the constant - multiple rule, if $y = 8\sin(x)$, then $y'=8\cos(x)$.

Step3: Differentiate $v(x)=x$

The derivative of $x$ with respect to $x$ is $1$.

Step4: Apply the difference - rule

$\frac{d}{dx}[8\sin(x)-x]=\frac{d}{dx}[8\sin(x)]-\frac{d}{dx}[x]=8\cos(x) - 1$.

Answer:

$8\cos(x)-1$