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onegirl

  • one year ago

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

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  1. ingenuus
    • one year ago
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    \[f(x)=(x^2+1)^3=(x^2+1)(x^2+1)(x^2+1)\] \[f'(x)=6x(x^2+1)^2\]

  2. ingenuus
    • one year ago
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    used only product rule

  3. onegirl
    • one year ago
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    is that using the chain rule?

  4. ingenuus
    • one year ago
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    no. i said i only used product rule.

  5. onegirl
    • one year ago
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    okay

  6. zepdrix
    • one year ago
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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.

  7. onegirl
    • one year ago
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    not 100%

  8. onegirl
    • one year ago
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    okay i'll try doing with the chain rule

  9. zepdrix
    • one year ago
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    Ok :)

  10. onegirl
    • one year ago
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    okay can i post it here and correct me if i i mistake somewhere

  11. zepdrix
    • one year ago
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    k

  12. onegirl
    • one year ago
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    \[d/dx ((x^2 + 1)^3) = 3(x^2 + 1)^2 d/dx(x^2 + 1) = 3(d/dx(x^2)(x^2 - 1)^2 = 3(2x)(x^2 = 1)^2 = 6x(x^2 + 1)^2 \]

  13. onegirl
    • one year ago
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    @zepdrix

  14. zepdrix
    • one year ago
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    I'm not quite sure why it changed from x^2+1 to x^2-1. Was that a typo?

  15. onegirl
    • one year ago
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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

  16. onegirl
    • one year ago
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    yes typo for that one

  17. zepdrix
    • one year ago
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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 :)

  18. onegirl
    • one year ago
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    thanks!

  19. zepdrix
    • one year ago
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    `without` chain rule is going to be a little tricky. They want you multiply out your binomial :O oh boy.

  20. onegirl
    • one year ago
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    ohh so the one @ingenuus showed is nto by without chain rule?

  21. onegirl
    • one year ago
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    not*

  22. onegirl
    • one year ago
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    the one ingenuus showed its using the product rule

  23. zepdrix
    • one year ago
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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

  24. onegirl
    • one year ago
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    Ohhhh okay got you

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