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ksaimouli

derivative

  • one year ago
  • one year ago

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  1. ksaimouli
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    |dw:1353885456771:dw|

    • one year ago
  2. ksaimouli
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    \[\ln(1-2x)^3\]

    • one year ago
  3. ksaimouli
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    @ivanmlerner

    • one year ago
  4. ivanmlerner
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    You need to use the chain rule here. ln is one function, something^3 another and 1-2x another.

    • one year ago
  5. ivanmlerner
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    Do you have any ideas on what the first step is?

    • one year ago
  6. ksaimouli
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    |dw:1353887083835:dw|

    • one year ago
  7. ivanmlerner
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    Almost, you forgot to derivate the function z^3. You did the derivative of the ln and the one inside the cubic power.

    • one year ago
  8. ksaimouli
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    so should i take derivative (1-2x)^3 i thought i should only take derivative of inside the ()^3

    • one year ago
  9. ksaimouli
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    |dw:1353888159532:dw|

    • one year ago
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    |dw:1353888193965:dw|

    • one year ago
  11. ivanmlerner
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    Now that is correct. Just to summarize you have the following:\[f(x)=\ln(x)\]\[g(x)=x^3\]\[h(x)=1-2x\]and you want\[\frac{df(g(h(x)))}{dx}=\frac{df}{dg}\frac{dg}{dh}\frac{dh}{dx}\]\[\frac{df}{dg}=\frac{1}{g}=\frac{1}{(1-2x)^3}\]\[\frac{dg}{dh}=3h^2=3(1-2x)^2\]\[\frac{dh}{dx}=-2\]

    • one year ago
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