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

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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

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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?
Are you wanting to simplify f'(x)?
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?

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Other answers:

they applied the chain rule, then factored, so the 3 in the denominator canceled.
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
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no, for product rule, it's not over d^2
ok, got it. Thanks
(fg)' = f ' g + g' f

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