# How do you find the derivative of f(x)=ln (x^2+2)?

Aug 24, 2016

Since this is a composite function, you need to use the chain rule

#### Explanation:

The chain rune says that the derivative of a composition of functions $f \left(x\right) = g \left(h \left(x\right)\right)$ is:

$f ' \left(x\right) = g ' \left(h \left(x\right)\right) \cdot h ' \left(x\right)$. But:

$g ' \left(h \left(x\right)\right) = \ln ' \left({x}^{2} + 2\right) = \frac{1}{{x}^{2} + 2}$, and

$h ' \left(x\right) = 2 x$, so the full derivative is:

f'(x)=1/(x^2+2)*2x=(2x)/(x^2+2

Aug 24, 2016

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

#### Explanation:

Let $y = f \left(x\right) = \ln \left({x}^{2} + 2\right)$

$\therefore {e}^{y} = {x}^{2} + 2. \ldots \ldots \ldots \ldots \left(\star\right)$.

$\Rightarrow \frac{d}{\mathrm{dx}} \left({e}^{y}\right) = \frac{d}{\mathrm{dx}} \left({x}^{2} + 2\right)$

But, by the Chain Rule, $\frac{d}{\mathrm{dx}} \left({e}^{y}\right) = \frac{d}{\mathrm{dy}} \left({e}^{y}\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$, and,

$\frac{d}{\mathrm{dx}} \left({x}^{2} + 2\right) = \frac{d}{\mathrm{dx}} \left({x}^{2}\right) + \frac{d}{\mathrm{dx}} \left(2\right) = 2 x + 0 = 2 x$.

Hence, ${e}^{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x$

$\therefore f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x}{e} ^ y = \frac{2 x}{{x}^{2} + 2}$.

Enjoy Maths.!