anonymous
  • anonymous
Find the derivative with and without using the chain rule. f(x) = (x^2 + 1)^3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[f(x)=(x^2+1)^3=(x^2+1)(x^2+1)(x^2+1)\] \[f'(x)=6x(x^2+1)^2\]
anonymous
  • anonymous
used only product rule
anonymous
  • anonymous
is that using the chain rule?

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anonymous
  • anonymous
no. i said i only used product rule.
anonymous
  • anonymous
okay
zepdrix
  • zepdrix
One, do you understand how to find the derivative `with` using chain rule? You would just take the derivative as is.
anonymous
  • anonymous
not 100%
anonymous
  • anonymous
okay i'll try doing with the chain rule
zepdrix
  • zepdrix
Ok :)
anonymous
  • anonymous
okay can i post it here and correct me if i i mistake somewhere
zepdrix
  • zepdrix
k
anonymous
  • anonymous
\[d/dx ((x^2 + 1)^3) = 3(x^2 + 1)^2 d/dx(x^2 + 1) = 3(d/dx(x^2)(x^2 - 1)^2 = 3(2x)(x^2 = 1)^2 = 6x(x^2 + 1)^2 \]
anonymous
  • anonymous
@zepdrix
zepdrix
  • zepdrix
I'm not quite sure why it changed from x^2+1 to x^2-1. Was that a typo?
anonymous
  • anonymous
i see that it didn't show up all of it but here is the last part 3(2x)(x^2 + 1)^2 = 6x(x^2 + 1)^2
anonymous
  • anonymous
yes typo for that one
zepdrix
  • zepdrix
\[\large f'(x)=3(x^2+1)^2\frac{d}{dx}(x^2+1)\]\[\large f'(x)=3(x^2+1)^2(2x) \qquad \qquad f'(x)=6x(x^2+1)^2\] Ok yah looks good :)
anonymous
  • anonymous
thanks!
zepdrix
  • zepdrix
`without` chain rule is going to be a little tricky. They want you multiply out your binomial :O oh boy.
anonymous
  • anonymous
ohh so the one @ingenuus showed is nto by without chain rule?
anonymous
  • anonymous
not*
anonymous
  • anonymous
the one ingenuus showed its using the product rule
zepdrix
  • zepdrix
That's another method for doing it without chain rule. Which involved the product rule with `3 terms`. Which is a little bit tricky if you've never seen it before. Product rule with 2 terms is probably what you're used to seeing. I was referring to a different method, multiplying out all the brackets as Wio showed you the last time you posted this. :D
anonymous
  • anonymous
Ohhhh okay got you

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