Question

If+y=logx/x,prove that d^2/dx^2=2logx-3/x^3

Answer 1

Answer :

$\frac{d^2y}{dx^2}=\frac{2logx-3}{x^3}$

Step-by-step explanation :

$y=\frac{logx}{x} [Since*use*u/v*method]\\\\=>\frac{dy}{dx}=\frac{ x\frac{1}{x}-logx*1}{x^2} \\\\ =>\frac{dy}{dx}=\frac{1-logx}{x^2} \\\\=>\frac{d^2y}{dx^2}=\frac{x^2.\frac{-1}{x}- (1-logx)2x}{x^4} \\\\=>\frac{d^2y}{dx^2}=\frac{-x-2x+2xlogx}{x^4}\\\\ =>\frac{d^2y}{dx^2}=\frac{-3x+2xlogx}{x^4}[Take*x*as*common]\\\\ =>\frac{d^2y}{dx^2}=\frac{2logx-3}{x^3}$

Hence Proved.

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