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anonymous
 one year ago
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)
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anonymous
 one year ago
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are you wanting to simplify f'(x)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0they applied the chain rule, then factored, so the 3 in the denominator canceled.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh, I think I figured it out f(x)*d/dx g(x) + g(x) * d/dx f(x) is all over d^2 right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442968002532:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no, for product rule, it's not over d^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(fg)' = f ' g + g' f
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