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

Feb 5, 2017

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

Explanation:

This can be rewritten as

$y = \ln {x}^{2} + \ln {e}^{x}$

We know that $y = \ln x$ and $y = {e}^{x}$ are inverses, such that $\ln e = 1$. Using the law $\ln {a}^{n} = n \ln a$, we can say that $\ln {e}^{x} = x$ and that $\ln {x}^{2} = 2 \ln x$.

$y = 2 \ln x + x$

$y ' = \frac{2}{x} + 1$

This is the first derivative. Differentiate this to get the second derivative.

$y ' = 2 {x}^{-} 1 + 1$

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

Hopefully this helps!