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anonymous
 3 years ago
Find the derivative with and without using the chain rule. f(x) = (x^2 + 1)^3
anonymous
 3 years ago
Find the derivative with and without using the chain rule. f(x) = (x^2 + 1)^3

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=(x^2+1)^3=(x^2+1)(x^2+1)(x^2+1)\] \[f'(x)=6x(x^2+1)^2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0used only product rule

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is that using the chain rule?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no. i said i only used product rule.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1One, do you understand how to find the derivative `with` using chain rule? You would just take the derivative as is.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay i'll try doing with the chain rule

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay can i post it here and correct me if i i mistake somewhere

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[d/dx ((x^2 + 1)^3) = 3(x^2 + 1)^2 d/dx(x^2 + 1) = 3(d/dx(x^2)(x^2  1)^2 = 3(2x)(x^2 = 1)^2 = 6x(x^2 + 1)^2 \]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1I'm not quite sure why it changed from x^2+1 to x^21. Was that a typo?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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
 3 years ago
Best ResponseYou've already chosen the best response.0yes typo for that one

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\large f'(x)=3(x^2+1)^2\frac{d}{dx}(x^2+1)\]\[\large f'(x)=3(x^2+1)^2(2x) \qquad \qquad f'(x)=6x(x^2+1)^2\] Ok yah looks good :)

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1`without` chain rule is going to be a little tricky. They want you multiply out your binomial :O oh boy.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ohh so the one @ingenuus showed is nto by without chain rule?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the one ingenuus showed its using the product rule

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1That'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
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