Question

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

Answer 1

`(x+2)/x`

Explanation:

This can be significantly simplified first using logarithm rules. The first we will use is that `log(AB)=log(A)+log(B)`, so:

`y=ln(x^2e^x)=ln(x^2)+ln(e^x)`

Since `ln(x)=log_e(x)` and `e^x`, are inverse functions, we see that `ln(e^x)=x`. Furthermore, we will also use `log(A^B)=Blog(A)` for `ln(x^2)`:

`y=2ln(x)+x`

Now, to differentiate this, we must know that `d/dxln(x)=1/x`. Thus:

`dy/dx=2/x+1=(x+2)/x`