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!
