# What is the second derivative of f(x)=ln (x+3)?

Apr 25, 2018

$f ' ' \left(x\right) = - \frac{1}{x + 3} ^ 2$

#### Explanation:

NOTE that the Derivative of $\ln \left(x\right) = \frac{1}{x}$ and derivative of $\frac{1}{x} = - \frac{1}{x} ^ 2$

$f ' \left(x\right) = \frac{d \left\{\ln \left(x + 3\right)\right\}}{\mathrm{dx}} = \frac{1}{x + 3}$

$f ' ' \left(x\right) = \frac{d \left\{\frac{1}{x + 3}\right\}}{\mathrm{dx}} = - \frac{1}{x + 3} ^ 2$

Apr 25, 2018

$f ' ' \left(x\right) = - \frac{1}{x + 3} ^ 2$

#### Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

$\text{Given "y=f(g(x))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$

$\Rightarrow f ' \left(x\right) = \frac{1}{x + 3} \times \frac{d}{\mathrm{dx}} \left(x + 3\right) = \frac{1}{x + 3} = {\left(x + 3\right)}^{-} 1$

$\text{differentiate "f'(x)" using the chain rule}$

$\Rightarrow f ' ' \left(x\right) = - {\left(x + 3\right)}^{-} 2 \times \frac{d}{\mathrm{dx}} \left(x + 3\right) = - \frac{1}{x + 3} ^ 2$