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anonymous

  • one year ago

the question is in the picture, why when i go from (1) or (a) i get different equation for the same g' ? https://dl.pushbulletusercontent.com/X0Sg1DgSq9wucZsrYb1KCdHUr2VWY2rJ/IMG_20150701_161911.jpg

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  1. anonymous
    • one year ago
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    This seems to be using u-substitution try this: https://www.khanacademy.org/math/integral-calculus/integration-techniques/u_substitution/v/u-substitution

  2. DanJS
    • one year ago
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    From a to b, did you divide by [g(x)]^4 ?

  3. anonymous
    • one year ago
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    no, doubled by g^4

  4. misty1212
    • one year ago
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    did you try it with an actual example?

  5. misty1212
    • one year ago
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    \[g(x)=\sin(x), f(x)=\frac{1}{\sin^3(x)}\]

  6. anonymous
    • one year ago
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    way to try with numbers? f(x) and g(x) can be any equations

  7. misty1212
    • one year ago
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    \[f'(x)=\frac{-3\cos(x)}{\sin^4(x)}\] and so

  8. misty1212
    • one year ago
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    \[g'(x)=-\frac{1}{3}f'(x)g^4(x)\] or \[\cos(x)=\cos(x)\]

  9. anonymous
    • one year ago
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    but i want to do implicit derivering, and stay with f(x) and g(x). are (3) and (c) the same?

  10. DanJS
    • one year ago
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    solve C for g' and check

  11. anonymous
    • one year ago
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    OK, got it, thanks! in (c): \[fg^3 \] = 1

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