Question

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

Answer 1

Let $\bf{y=log(logx)}$
now differentiate y with respect to x,
$\bf{\frac{dy}{dx}=\frac{d\{log(logx)\}}{dx}}\\\\=\bf{\frac{1}{logx}\frac{d(logx)}{dx}}\\\\=\bf{\frac{1}{x.logx}}$

so, $\bf{\frac{dy}{dx}=\frac{1}{x.logx}}$

now differentiate $\bf{\frac{dy}{dx}}$ with respect to x once again,
$\bf{\frac{d^2y}{dx^2}=\frac{d\{(x.logx)^{-1}\}}{dx}}\\\\=\bf{-(x.logx)^{-1-1}\frac{d(x.logx)}{dx}}\\\\=\bf{\frac{-1}{(x.logx)^2}\left[\begin{array}{c}x.\frac{d(logx)}{dx}+logx\frac{dx}{dx}\end{array}\right]}\\\\=\bf{\frac{-1}{(x.logx)^2}[1+logx]}$

hence, d²y/dx² = -(1 + logx)/(x.logx)²

Answer 2

let y = log(logx)

so, dy/dx = 1/(logx) * d(logx)/dx

dy/dx = 1/(xlogx)

now,

d²y/dx² = d{1/(xlogx)}/dx

using division rule,

d²y/dx² = $\mathit{\frac{xlogx * 0 - 1*(x*1/x + logx*1)}{(xlogx)^2}}$

d²y/dx² = $\mathit{\frac{0 - (1 + logx)}{(xlogx)^2}}$

d²y/dx² = $\mathit{\frac{-(1 + logx)}{(xlogx)^2}}$