# What is the second derivative of y = lnx^2?

Aug 1, 2016

For the first derivative, use the chain rule.

$y = \ln \left(u\right)$ and $u = {x}^{2}$. Then $y ' = \frac{1}{u}$ and $u ' = 2 x$.

$y ' = 2 x \times \frac{1}{u} = 2 x \times \frac{1}{{x}^{2}} = \frac{2 x}{x} ^ 2$

So, the first derivative is $\frac{2 x}{x} ^ 2$. The second derivative can be determined by differentiating the first.

By the quotient rule:

Let $y = \frac{g \left(x\right)}{h \left(x\right)}$, so that $g \left(x\right) = 2 x$ and $h \left(x\right) = {x}^{2}$. The derivative is given by $y ' = \frac{g ' \left(x\right) \times h \left(x\right) - g \left(x\right) \times h ' \left(x\right)}{h \left(x\right)} ^ 2$

The derivative of $g \left(x\right)$ is $2$ and the derivative of $h \left(x\right)$ is $2 x$.

We can now substitute into the formula and calculate. Note: the notation $y ' '$ is used to show that we're finding the second derivative, and not the first, as would the notation $y '$. Similarly, $y ' ' '$ would signify the third derivative.

$y ' ' = \frac{2 \times {x}^{2} - 2 x \times 2 x}{{x}^{2}} ^ 2$

$y ' ' = \frac{2 {x}^{2} - 4 {x}^{2}}{x} ^ 4$

$y ' ' = \frac{- 2 {x}^{2}}{x} ^ 4$

$y ' ' = - \frac{2}{x} ^ 2$

Hence, the second derivative is $y ' ' = - \frac{2}{x} ^ 2$.

Hopefully this helps!