Question

I have a calculus problem:
cubed root of (x^3+6x+1) times x^7

I have worked out part of this problem but there is a part which I will identify that I am unsure of how the solution is arrived at.

Let F(x)=f(x)*g(x)

Let f(x)= (x³+6x+1)^(1/3)
d/dx (x³+6x+1)^(1/3)= (x²+2)/(x³+6x+1)^(2/3)
Let g(x)=x^7
d/dx x^7= 7x^6

F'(x)= (x³+6x+1)^(1/3)*7x^6 +x^7 * (x²+2)/(x³+6x+1)^(2/3)

read next

Answer 1

Now here is where I am having problems . . .
The next step in the solution shows the f(x) as [(x³+6x+x)^(1/3) * 7x^6]/(x³+6x+1)^(2/3) + a second portion which I will leave off for the time being. My question is how [(x³+6x+x)^(1/3) got in the numerator and what happened to the 3 that should be in the denominator?

Answer 2

Are you wanting to simplify f'(x)?

Answer 3

I am trying to figure out what happened with f(x). When I brought the 1/3 in front what happened to to 3 that should be in the denominator?

and when I took the (x^3+6x+1)^(-2/3) down to the denominator why did (X^3+6x+1) remain in the numerator?

Answer 4

they applied the chain rule, then factored, so the 3 in the denominator canceled.

Answer 5

oh, I think I figured it out
f(x)*d/dx g(x) + g(x) * d/dx f(x) is all over d^2 right