Question

if ( y=cos x-ln (2 x) ), then ( \frac{d^{2} y}{d x^{2}}= ) ( sin x-\frac{2}{x^{2}} ) ( -sin x-\frac{2}{x^{2}} ) ( sin x-\frac{1}{x^{2}} ) ( -sin x-\frac{1}{x^{2}} )

Explanation:

Step1: Differentiate \(y = \cos x-\ln(2x)\) once

Use the derivative rules: \((\cos x)'=-\sin x\), \((\ln u)'=\frac{u'}{u}\) (where \(u = 2x\), \(u'=2\)).
So \(y'=-\sin x-\frac{2}{2x}=-\sin x-\frac{1}{x}\).

Step2: Differentiate \(y'\) again

Differentiate \(-\sin x\) and \(-\frac{1}{x}\) separately. \((-\sin x)'=-\cos x\)'s second - derivative: \((-\sin x)'=-\cos x\)'s derivative is \(\sin x\); \((-\frac{1}{x})'=-x^{-1}\)'s derivative is \(x^{-2}=\frac{1}{x^{2}}\) (using the power rule \((x^n)'=nx^{n - 1}\)).
So \(y''=\sin x-\frac{1}{x^{2}}\).

Answer:

C. \(\sin x-\frac{1}{x^{2}}\)