Find the second derivative of $e^{x^{3}}+7x$

Question

$$e^{x^{3}}+7x$$ 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.

Answer 1

$$\frac d{dx}\left(e^{x^3}\right) = e^{x^3}\cdot \bigl(x^3\bigr)' = e^{x^3}\bigl(3x^2\bigr) = 3e^{x^3}x^2$$

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 $$\frac {d^2}{dx}\left(e^{x^3}\right) = \frac{d}{dx}\Bigl(3e^{x^3}x^2\Bigr) = \Bigl(3e^{x^3}\Bigr)\Bigl(x^2\Bigr)' + x^2\Bigl(3e^{x^3}\Bigr)' = 6xe^{x^3} + 9x^4e^{x^3}$$

Answer 2

Hint:

$$\left(e^{x^3}\right)'=3x^2e^{x^3}\;,\;\;\left(3x^2e^{x^3}\right)'=6xe^{x^3}+9x^4e^{x^3}=3xe^{x^3}\left(2+3x^3\right)$$

I think you can now handle the summand $\,7x\,$

Answer 3

Let

$ f(x) = e^x $ and $ g(x) = x^3 $, so that your first term could be written $ f(g(x)) $. Then the derivative of the first term is

$ f'(g(x)) g'(x) = e^{g(x)} (3x^2) = e^{x^3} 3x^2 $

Answer 4

let T be the second derivative of the function ,

then ,

T = $ 3 e^{x^3} [ 3x^4 + 2x ] $