Question

find the derivative of $f(x) = -5x^2log_7(2x^4 + 13)$.\
\
$\boldsymbol{f(x) = -\dfrac{80x^4}{(2x^4 + 13)\ln 7}}$\
\
$\boldsymbol{f(x) = -10x\log_7(2x^4 + 13) - \dfrac{40x^5}{2x^4 + 13}}$\
\
$\boldsymbol{f(x) = -10x\log_7(2x^4 + 13) - \dfrac{40x^5}{(2x^4 + 13)\ln 7}}$\
\
$\boldsymbol{f(x) = -\dfrac{80x^4}{2x^4 + 13}}$

Explanation:

Step1: Recall product rule

Let $u(x) = -5x^2$, $v(x)=\log_7(2x^4+13)$. Product rule: $f'(x)=u'(x)v(x)+u(x)v'(x)$

Step2: Compute $u'(x)$

Derivative of $-5x^2$: $u'(x) = -10x$

Step3: Convert log to natural log

$\log_7(2x^4+13)=\frac{\ln(2x^4+13)}{\ln7}$

Step4: Compute $v'(x)$

Use chain rule: $v'(x)=\frac{1}{\ln7}\cdot\frac{8x^3}{2x^4+13}=\frac{8x^3}{(2x^4+13)\ln7}$

Step5: Substitute into product rule

$f'(x)=-10x\log_7(2x^4+13) + (-5x^2)\cdot\frac{8x^3}{(2x^4+13)\ln7}$
Simplify the second term: $(-5x^2)\cdot\frac{8x^3}{(2x^4+13)\ln7}=-\frac{40x^5}{(2x^4+13)\ln7}$

Answer:

$f'(x)= - 10x\log_{7}(2x^{4}+13)-\frac{40x^{5}}{(2x^{4}+13)\ln7}$ (the third option)