General formula for a second derivative

I'm struggling trying to find the second derivative of

f(x)=x^4*e^x

I've found f ' (x)= e^x*x^3(4+x) using the product rule.

I'm not very good with the dy/dx notations yet and have been using u, v, u' and v' when doing my equations, so f ' (x) = [(u' * v) + (v' * u)] / (v)^2

So:
u=x^4
u'=4x^3
v=e^x
v'=e^x

I understand that I will need to get more familiar with the dy/dx format, but I'm not quite there and am looking for a simple formal to calculate f '' (x) in terms of u and v.

Thank you!!

What you have for f' is correct, but you shouldn't have pulled out the factor and simplified. This is because it will be easier to find f" if you leave it:
f' = x^4*e^x + e^x*4x^3. Now find f" of each term separately.
Use the product rule on x^4*e^x,
and then use the product ruleon e^x*4x^3.
Then, you may want to pull out a common factor as this will be your final answer.

You seem to know what you are doing, so i think you will be fine from here.
email w/questions

p.s. [(u' * v) + (v' * u)] / (v)^2 is the quotient rule. oops!!

dy/dx means the derivative of the y function (f(x), with respect to x.
Basically, this simply means x' = 1
Soon you will do implicit differentiation with respect to x.
dy/dx using implicit differentiation means the derivative of X^2 = 2x,
while the derivative of y^2 is 2yy', because in dy/dx: y' = y', not 1
your teacher will explain when the time comes

Additional Info:
This stuff about dy/dx is for future use, not this problem.
Hope I didn't confuse?

I would have rated you both best if I could lol I solved the second derivative using the ways you both suggested