what is the derivative of \\( \\ln ( x ) - 8 \\cos ( x ) \\)?\n\\( \\frac { 1 } { x } - \\sin ( 8 x )…

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