let (f(x)= - 8ln(2x^{2})). find (f(x)) and (f(-2)).\n\n(f(x)=)\n\n(f(-2)=)
let (f(x)= - 8ln(2x^{2})). find (f(x)) and (f(-2)).\n\n(f(x)=)\n\n(f(-2)=)
Answer
Explanation:
Step1: Apply chain - rule and derivative of ln(u)
The derivative of $\ln(u)$ is $\frac{u'}{u}$. Let $u = 2x^{2}$, then $u'=4x$. The function $y=-8\ln(2x^{2})$. By the constant - multiple rule and chain - rule, $y'=-8\times\frac{4x}{2x^{2}}$.
Step2: Simplify the derivative expression
Simplify $-8\times\frac{4x}{2x^{2}}$. First, $-8\times\frac{4x}{2x^{2}}=\frac{-32x}{2x^{2}}=-\frac{16}{x}$. So $f'(x)=-\frac{16}{x}$.
Step3: Evaluate $f'(-2)$
Substitute $x = - 2$ into $f'(x)$. $f'(-2)=-\frac{16}{-2}=8$.
Answer:
$f'(x)=-\frac{16}{x}$ $f'(-2)=8$