onegirl
Find the derivative with and without using the chain rule. f(x) = (x^2 + 1)^3
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ingenuus
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\[f(x)=(x^2+1)^3=(x^2+1)(x^2+1)(x^2+1)\]
\[f'(x)=6x(x^2+1)^2\]
ingenuus
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used only product rule
onegirl
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is that using the chain rule?
ingenuus
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no. i said i only used product rule.
onegirl
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okay
zepdrix
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One, do you understand how to find the derivative `with` using chain rule?
You would just take the derivative as is.
onegirl
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not 100%
onegirl
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okay i'll try doing with the chain rule
zepdrix
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Ok :)
onegirl
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okay can i post it here and correct me if i i mistake somewhere
zepdrix
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k
onegirl
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\[d/dx ((x^2 + 1)^3) = 3(x^2 + 1)^2 d/dx(x^2 + 1) = 3(d/dx(x^2)(x^2 - 1)^2 = 3(2x)(x^2 = 1)^2 = 6x(x^2 + 1)^2 \]
onegirl
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@zepdrix
zepdrix
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I'm not quite sure why it changed from x^2+1 to x^2-1. Was that a typo?
onegirl
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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
onegirl
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yes typo for that one
zepdrix
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\[\large f'(x)=3(x^2+1)^2\frac{d}{dx}(x^2+1)\]\[\large f'(x)=3(x^2+1)^2(2x) \qquad \qquad f'(x)=6x(x^2+1)^2\]
Ok yah looks good :)
onegirl
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thanks!
zepdrix
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`without` chain rule is going to be a little tricky. They want you multiply out your binomial :O oh boy.
onegirl
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ohh so the one @ingenuus showed is nto by without chain rule?
onegirl
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not*
onegirl
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the one ingenuus showed its using the product rule
zepdrix
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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
onegirl
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Ohhhh okay got you