Question

what is the derivative of \\( \ln ( x ) - 8 \cos ( x ) \\)?
\\( \frac { 1 } { x } - \sin ( 8 x ) \\)
\\( \frac { 1 } { x } + \sin ( 8 x ) \\)
\\( x \ln ( x ) - 8 \sin ( x ) \\)
\\( x \ln ( x ) + 8 \sin ( x ) \\)
\\( \frac { 1 } { x } - 8 \sin ( x ) \\)
\\( \frac { 1 } { x } + 8 \sin ( x ) \\)
clear my selection

Explanation:

Step1: Differentiate $\ln(x)$

The derivative of $\ln(x)$ is $\frac{1}{x}$.

Step2: Differentiate $-8\cos(x)$

Using the rule that the derivative of $\cos(x)$ is $-\sin(x)$, for $y = -8\cos(x)$, by the constant - multiple rule $(cf(x))^\prime=c f^\prime(x)$ (where $c = - 8$ and $f(x)=\cos(x)$), we have $(-8\cos(x))^\prime=-8\times(-\sin(x)) = 8\sin(x)$.

Step3: Sum the derivatives

By the sum rule $(u + v)^\prime=u^\prime+v^\prime$ (here $u=\ln(x)$ and $v=-8\cos(x)$), the derivative of $\ln(x)-8\cos(x)$ is $\frac{1}{x}+8\sin(x)$.

Answer:

$\frac{1}{x}+8\sin(x)$ (the last option)