Question

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

Answer 1

Let $\bf{y=sin(logx)}$
now differentiate y with respect to x
$\bf{\frac{dy}{dx}=\frac{d[sin(logx)]}{dx}}\\\\=\bf{cos(logx)\frac{d(logx)}{dx}}\\\\=\bf{cos(logx).\frac{1}{x}}\\\\=\bf{\frac{cos(logx)}{x}}$

so, $\bf{\frac{dy}{dx}=\frac{cos(logx)}{x}}$

differentiate $\frac{dy}{dx}$ with respect to x once again,

$\bf{\frac{d^2y}{dx^2}=\frac{d\left[\begin{array}{c}\frac{cos(logx)}{x}\end{array}\right]}{dx}}\\\\=\bf{\frac{x.\frac{d(cos(logx)}{dx}-cos(logx).\frac{dx}{dx}}{x^2}}\\\\=\bf{\frac{-sin(logx)-cos(logx)}{x^2}}$

hence, d²y/dx² =- {sin(logx) + cos(logx)}/x²