Question

Let $g\left(x\right)=x^3\ln \left(x^{2}f\right(x\left)\right)$, where $f\left(x\right)$ is a differentiable, positive function on $(0,\infty )$ satisfying $f\left(2\right)=\frac{1}{4}$ and $f^{\prime }\left(2\right)=-3$, then $g^{\prime }\left(2\right)$ equal to

• A. $77$
• B. $-88$
• C. $88$
• D. $-77$

Answer 1

Answer: (B)

$-88$

$g\left(x\right)=x^3\ln \left(x^{2}f\right(x\left)\right)$
Differentiating both sides w.r.t. $x$, we get
$g^{\prime }\left(x\right)=x^3\frac{x^2{f}^{\prime }\left(x\right)+2xf\left(x\right)}{x^{2}f\left(x\right)}+3x^2\ln \left(x^{2}f\right(x\left)\right)$
$g^{\prime }\left(2\right)=8\frac{4f^{\prime }\left(x\right)+4f\left(x\right)}{4f\left(x\right)}+12\ln \left(4f\right(x\left)\right)$

$g^{\prime }\left(2\right)=8\times \frac{-12+4\times \frac{1}{4}}{1}+12\ln (4\times \frac{1}{4})=-88$