What is the derivative of $f(x) = e^(x^2 + lnx^3)$ ?

Question

1224 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-18T12:15:01

$f'(x) = e^(x^2+lnx^3) (2x + 3/x)$

We have:

$f(x) = e^(x^2+lnx^3)$

So by the chain rule we have:

$f'(x) = e^(x^2+lnx^3) \ d/dx {x^2+lnx^3}$

$\ \ \ \ \ \ \ \ = e^(x^2+lnx^3) d/dx {x^2 + 3lnx}$

$\ \ \ \ \ \ \ \ = e^(x^2+lnx^3) (2x + 3/x)$

What is the derivative of f(x) = e^(x^2 + lnx^3)f(x) = e^(x^2 + lnx^3)?