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

ingenuus
 one year ago
Best ResponseYou've already chosen the best response.0used only product rule

onegirl
 one year ago
Best ResponseYou've already chosen the best response.0is that using the chain rule?

ingenuus
 one year ago
Best ResponseYou've already chosen the best response.0no. i said i only used product rule.

zepdrix
 one year 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.

onegirl
 one year ago
Best ResponseYou've already chosen the best response.0okay i'll try doing with the chain rule

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

onegirl
 one year 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
 one year 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?

onegirl
 one year 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

zepdrix
 one year 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
 one year 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.

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

onegirl
 one year ago
Best ResponseYou've already chosen the best response.0the one ingenuus showed its using the product rule

zepdrix
 one year 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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