# How do you find f'(x) and f''(x) given f(x)=x^4e^x?

Nov 25, 2016

Use the product rule to take the first derivative, than again to take the second. This yields $f ' \left(x\right) = 4 {x}^{3} {e}^{x} + {x}^{4} {e}^{x}$ and $f ' ' \left(x\right) = {x}^{4} {e}^{x} + 8 {x}^{3} {e}^{x} + 12 {x}^{2} {e}^{x}$.

#### Explanation:

By the product rule, for a function $f g$, the derivative is given by $f ' g + f g '$ or $f g ' + f ' g$. The order does not matter because addition is commutative.

To find the first derivative of the function, $f ' \left(x\right)$, first take the derivative of ${x}^{4}$ and multiply this by ${e}^{x}$, which we have left unchanged. This gives:

$4 {x}^{3} \cdot {e}^{x}$

Now, take the derivative of ${e}^{x}$ (which is still ${e}^{x}$ by properties of exponential functions) and multiply this by ${x}^{4}$, which we have left unchanged. This gives:

${e}^{x} \cdot {x}^{4}$

$f ' \left(x\right) = 4 {x}^{3} {e}^{x} + {x}^{4} {e}^{x}$

Finding the second derivative follows a similar process. Our starting function is now $f ' \left(x\right)$. We will use the product rule to take the derivative of each "half" that we found above for $f ' \left(x\right)$, and add those halves together (as specified in the function).

First, we take the derivative of $4 {x}^{3} {e}^{x}$. Using the product rule, we take the derivative of $4 {x}^{3}$ and multiply this by ${e}^{x}$, which we leave unchanged, then add this to the derivative of ${e}^{x}$, which we multiply by $4 {x}^{3}$. This gives:

$12 {x}^{2} {e}^{x} + 4 {x}^{3} {e}^{x}$

Now, we take the derivative of ${x}^{4} {e}^{x}$ following the same process. We get:

$4 {x}^{3} {e}^{x} + {x}^{4} {e}^{x}$

Now, we add these two halves together to get:

$f ' ' \left(x\right) = 12 {x}^{2} {e}^{x} + 4 {x}^{3} {e}^{x} + 4 {x}^{3} {e}^{x} + {x}^{4} {e}^{x}$

This simplifies to:

$f ' ' \left(x\right) = {x}^{4} {e}^{x} + 8 {x}^{3} {e}^{x} + 12 {x}^{2} {e}^{x}$

or further to:

$f ' ' \left(x\right) = {e}^{x} \left({x}^{4} + 8 {x}^{3} + 12 {x}^{2}\right)$

Hope that helps!