Question

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

Answer 1

`y' = (2 + x)/x`
`y'' = -2/x^2`

Explanation:

This can be rewritten as

`y = lnx^2 + lne^x`

We know that `y= lnx` and `y = e^x` are inverses, such that `lne = 1`. Using the law `lna^n = nlna`, we can say that `lne^x = x` and that `lnx^2 = 2lnx`.

`y = 2lnx + x`

`y' = 2/x + 1`

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

`y' = 2x^-1 + 1`

`y'' = -2/x^2`

Hopefully this helps!