Chain rule confusion

[MathNoob94]

Hi, I have this question im working on and I can't seem to figure out what to do. The question is f(x) =(4x-x^2)^3 what I have is f'(x) =3(4x-2)^2 times (4-2x) do I leave my final answer in this form or will I need to do some simplifying? If so please elaborate on the steps as I can't figure out how to do so.

Thank you!!

 

[stapel]

The question is f(x) =(4x-x^2)^3 what I have is f'(x) =3(4x-2)^2 times (4-2x)

If "the question is f(x) = (4x - x^2)^3", with no instructions, then there is nothing to do, as you have not been told what is required to be done.

If "what [you] have is f'(x)...", then I will guess that there were instructions, and those instructions said something along the lines of "find the derivative with respect to x".

do I leave my final answer in this form or will I need to do some simplifying?

The answer to that will lie in the instructions (do they say to "simplify"? if so, how does your particular book interpret the term "simplify" in this context?) and your instructor or grader.

If so please elaborate on the steps as I can't figure out how to do so.

Use what you learned back in algebra and pre-calc: multiply out the polynomial terms.

 

[MathNoob94]

Sorry it I must use the chain rule

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[stapel]

Sorry it I must use the chain rule

...f(x) =(4x-x^2)^3 what I have is f'(x) =3(4x-2)^2 times (4-2x)

I see where you applied the Chain Rule in differentiating the third power:

. . . . .\(\displaystyle f(x)\, =\, g(h(x))\, =\, (4x\, -\, x^2)^3\)

. . . . .\(\displaystyle \mbox{where }\, h(x)\, =\, 4x\, -\,x^2\, \mbox{ and }\, g(x)\, =\, x^3\)

However, when you differentiate the outer function (what I've called "g(x)" above), you should differentiate ONLY that outer function; the insides stay the same. So there is no way that the argument inside should go from 4x - x² to 4x - 2. What happened there?

 

[MathNoob94]

I followed f'(g(x)) × g'(x)

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[MathNoob94]

I followed f'(g(x)) × g'(x)

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So I just did the derivative of g of x

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[stapel]

I followed f'(g(x)) × g'(x)

Okay; I'll guess that you've renamed "f(x) = g(h(x))" from my post (and the original exercise) to something like "h(x) = f(g(x))", with h(x) taking the place of the original function name of f(x).

However, this does not answer how you took g(x) = 4x - x², applied the Chain Rule to the outer function of f(x) = x³, and arrived at f'(g(x)) = f'(4x - 2)*g'(x). Your argument, 4x - 2, does not equal your g(x) (nor does it equal the derivative of g(x)).

Please reply showing all of your steps and clearly stating your reasoning. Thank you.