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

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}} )

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}} )

Answer

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}})