How do you find the first and second derivative of lnx^2 ?

Oct 10, 2016

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

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

Explanation:

Using the chain rule we get:

$y ' = \left(\frac{1}{x} ^ 2\right) \cdot 2 x$=$2 \frac{x}{x} ^ 2$

$y ' ' = \frac{\left({x}^{2} \cdot 2\right) - 4 {x}^{2}}{x} ^ 4$=$\frac{2 {x}^{2} - 4 {x}^{2}}{x} ^ 4$=$- 2 {x}^{2} / {x}^{4}$=$\frac{2}{x} ^ 2$