Question

if y = cos x - ln(2x), then $\frac{d^{3}y}{dx^{3}}=$
a sin x - $\frac{2}{x^{3}}$
b - sin x - $\frac{2}{x^{3}}$
c sin x - $\frac{1}{x^{3}}$
d - sin x - $\frac{1}{x^{3}}$

Explanation:

Step1: Recall derivative rules

The derivative of $\cos x$ is $-\sin x$ and the derivative of $\ln(u)$ with respect to $x$ is $\frac{u'}{u}$ (chain - rule). For $u = 2x$, $u'=2$, so the derivative of $\ln(2x)$ is $\frac{2}{2x}=\frac{1}{x}$. The first - derivative of $y=\cos x-\ln(2x)$ is $y'=-\sin x-\frac{1}{x}$.

Step2: Find the second - derivative

The derivative of $-\sin x$ is $-\cos x$ and the derivative of $-\frac{1}{x}=-x^{-1}$ using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$ is $x^{-2}=\frac{1}{x^{2}}$. So $y''=-\cos x+\frac{1}{x^{2}}$.

Step3: Find the third - derivative

The derivative of $-\cos x$ is $\sin x$ and the derivative of $\frac{1}{x^{2}}=x^{-2}$ using the power - rule is $-2x^{-3}=-\frac{2}{x^{3}}$. So $\frac{d^{3}y}{dx^{3}}=\sin x-\frac{2}{x^{3}}$.

Answer:

A. $\sin x-\frac{2}{x^{3}}$