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

1 Answer
Nov 25, 2016

Answer:

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

Explanation:

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

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

#4x^3*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*x^4#

Adding these two halves together, we get a final answer of:

#f'(x)=4x^3e^x+x^4e^x#

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

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

#12x^2e^x+4x^3e^x#

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

#4x^3e^x+x^4e^x#

Now, we add these two halves together to get:

#f''(x)=12x^2e^x+4x^3e^x+4x^3e^x+x^4e^x#

This simplifies to:

#f''(x)=x^4e^x+8x^3e^x+12x^2e^x#

or further to:

#f''(x)=e^x(x^4+8x^3+12x^2)#

Hope that helps!