# How do you find the first and second derivative of  {ln(x)}^2?

Jun 30, 2016

Let your function be $f \left(x\right) = {\left(\ln x\right)}^{2}$, then by the chain rule, we have:

$f ' \left(x\right) = \frac{1}{x} \times 2 \ln x = \frac{2 \ln \left(x\right)}{x}$

This is our first derivative. We obtain the second derivative by differentiating the first derivative.

If $f ' \left(x\right) = g \frac{x}{h \left(x\right)}$, then $f ' ' \left(x\right) = \frac{g ' \left(x\right) \times h \left(x\right) - g \left(x\right) \times h ' \left(x\right)}{h \left(x\right)} ^ 2$

$g \left(x\right) = 2 \ln x \to g ' \left(x\right) = \frac{2}{x}$

$h \left(x\right) = x \to h ' \left(x\right) = 1$

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

$f ' ' \left(x\right) = \frac{2 - 2 \ln x}{x} ^ 2 \text{or} \frac{2 \left(1 - \ln x\right)}{x} ^ 2$

In summary:

The first derivative of ${\left\{\ln \left(x\right)\right\}}^{2}$ is $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \ln \left(x\right)}{x}$

The second derivative of ${\left\{\ln \left(x\right)\right\}}^{2}$ is $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{2 \left(1 - \ln x\right)}{x} ^ 2$.

Practice exercises:

1. Determine the first, and second derivatives, if they exist.

a) $y = {\left(x - 2\right)}^{- 1}$

b) $y = 5 {x}^{5} - 7 {x}^{4} + 3 {x}^{3} - 9 {x}^{2} + 2$

c) $f \left(x\right) = {e}^{15 x} \times \ln {x}^{3}$

d) ${\left(\ln \left({\sin}^{2} x\right)\right)}^{3}$

Bonus:

Determine the third derivatives of the functions above, if they exist.

Good luck, and hopefully this helps!