Question

if (y = cos x-ln(2x)), then (\frac{d^{3}y}{dx^{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), where $u = 2x$ and $u'=2$. So the first - derivative of $y=\cos x-\ln(2x)$ is $y'=-\sin x-\frac{2}{2x}=-\sin x - \frac{1}{x}$.

Step2: Find the second - derivative

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

Answer:

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