$$e^{x^{3}}+7x$$
Here is what I have tried so far,
$$y'=(e^{x^{3}})'+(7x)'$$
$$y'=3e^{x^{2}}e^x+7$$
$$y'=3e^{x^{3}+x}+7$$
I am supposed to find the second derivative but I believe I have already made a mistake. I am not sure how I would apply the chain rule to the $e^{x^{3}}$. I checked Wolframalpha and the derivative of $e^{x^{3}}$ was $$3e^{x^{3}}x^2$$ Did I leave out a step when I applied chain rule? I went over the problem a couple of times and couldn't catch my mistake.
Best Answer
$$\frac d{dx}\left(e^{x^3}\right) = e^{x^3}\cdot \bigl(x^3\bigr)' = e^{x^3}\bigl(3x^2\bigr) = 3e^{x^3}x^2$$
This is by the chain rule, and for the exponential function, this gives us $$\frac d{dx}\left(e^{f(x)}\right) = e^{f(x)}f'x$$
Then for the second derivative, we can use the product rule (and the chain rule again):
So $$\frac {d^2}{dx}\left(e^{x^3}\right) = \frac{d}{dx}\Bigl(3e^{x^3}x^2\Bigr) = \Bigl(3e^{x^3}\Bigr)\Bigl(x^2\Bigr)' + x^2\Bigl(3e^{x^3}\Bigr)' = 6xe^{x^3} + 9x^4e^{x^3}$$