Question

Find the second order derivatives of the function
$sin(logx)$

Answer 1

The correct answer is $=\frac{-[sin(logx)+cos(logx)]}{x^2}$
Let $y=sin(logx)$
Then,
$\frac{dy}{dx}=\frac{d}{dx}(sin(logx))=cos(logx).\frac{d}{dx}(logx)=\frac{cos(logx)}{x}$
$∴\frac{d^2y}{dx^2}=\frac{d}{dx}[\frac{cos(logx)}{x}]$
$=\frac{x.\frac{d}{dx}[cos(logx)]-cos(logx).\frac{d}{dx}(x)}{x^2}$
$=\frac{x.[-sin(logx).\frac{d}{dx}(logx)]-cos(logx).1}{x^2}$
$=\frac{-xsin(logx).\frac{1}{x}-cos(logx)}{x^2}$
$=\frac{-[sin(logx)+cos(logx)]}{x^2}$