If $y=logx^5$, then $\frac{d^2y}{dx^2}$ is given by :

$\frac{1}{x^5}$ $\frac{1}{5x^5}$ $-\frac{20}{x^2}$ $-\frac{5}{x^2}$

$-\frac{5}{x^2}$

The correct answer is Option (4) → $-\frac{5}{x^2}$ $y=\log x^5=5\log x$ so $\frac{dy}{dx}=\frac{5}{x}$ so $\frac{d^2y}{dx^2}=-\frac{5}{x^2}$