What is the second derivative of #f(x)= e^(x^3)#?

Question

Charles Arocha · Accepted Answer

#f''(x)=3xe^(x^3)(3x^3+2)#

The first derivative (using the chain rule):

#f'(x)=d/dx[e^(x^3)]=e^(x^3)d/dx[x^3]=3x^2e^(x^3)#

Now, use the product rule as well to find the second derivative:

#f''(x)=3x^2d/dx[e^(x^3)]+e^(x^3)d/dx[3x^2]#

#f''(x)=3x^2(3x^2e^(x^3))+e^(x^3)(6x)#

#f''(x)=3xe^(x^3)(3x^3+2)#

Emily Ammerman · Answer

undefined To find the second derivative of $ f(x) = e^{x^3} $, we first need to find its first derivative and then differentiate that result with respect to $ x $ again.

Given $ f(x) = e^{x^3} $, the first derivative is:

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

Now, to find the second derivative, we differentiate $ f'(x) $ with respect to $ x $:

$$ f''(x) = (e^{x^3} \cdot 3x^2)' = (e^{x^3})' \cdot 3x^2 + e^{x^3} \cdot (3x^2)' $$

$$ = e^{x^3} \cdot (x^3)' \cdot 3x^2 + e^{x^3} \cdot 3 \cdot (x^2)' $$

$$ = e^{x^3} \cdot 3x^2 \cdot 3x^2 + e^{x^3} \cdot 3 \cdot 2x $$

$$ = 9x^4 e^{x^3} + 6x e^{x^3} $$

So, the second derivative of $ f(x) = e^{x^3} $ is $ f''(x) = 9x^4 e^{x^3} + 6x e^{x^3} $.