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Math Help - chain rule + differentiation

  1. #1
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    chain rule + differentiation

    missed a calc lesson and trying to play catch up. a little confused on this one:

    find the derivative of the function using chain rule or logarithmic differentiation

    f(x) = (x^2 + 2)^200 * (x^3 - x)^10

    any step by step help is greatly appreciated
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  2. #2
    Super Member craig's Avatar
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    Quote Originally Posted by JGaraffa View Post
    missed a calc lesson and trying to play catch up. a little confused on this one:

    find the derivative of the function using chain rule or logarithmic differentiation

    f(x) = (x^2 + 2)^200 * (x^3 - x)^10

    any step by step help is greatly appreciated
    Two solve this I'd use two facts.

    If u = u(x) and v = v(x), then \frac{d}{dx}(uv) = v.du + u.dv, (product rule)

    And the for the chain rule.

    \frac{d}{dx}(f(x))^n = n.f'(x).(f(x))^{n-1}.

    This is all you need to solve your question. Post all your working and we can help if you get stuck.
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  3. #3
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    Hello, JGaraffa!

    I'll do it both ways for you . . .



    Differentiate using the chain rule or log differentiation:

    . . f(x) \:=\:(x^2 + 2)^{200}(x^3 - x)^{10}

    Product and Chain Rules:


    . . f'(x) \;=\;(x^2+2)^{200}\cdot10(x^3-x)^0(3x^2-1) + (x^3-x)^{10}\cdot200(x^2+2)^{199}2x


    Factor: . f'(x) \;=\;10(x^2+2)^{199}(x^3-x)^9\,\bigg[(x^2+2)(3x^2-1) + (40x(x^3-x)\bigg]

    . . . . . . . . . =\;10(x^2+2)^{199}(x^3-x)^9\bigg[3x^4 + 6x^2 - x^2 - 2 + 40x^4 - 40x^2\bigg]

    . . . . . . . . . =\;10(x^2+2)^{199}(x^3-x)^9\left(43x^4 - 35x^2 - 2\right]


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Logarithmic Differentiaton:


    Take logs:

    . . \ln(y) \;=\;\ln\bigg[(x^2+2)^{200}(x^3-x)^{10}\bigg]

    . . . . . . =\;\ln(x^2+2)^{200} + \ln(x^3-x)^{10}

    . . . . . . =\;200\ln(x^2+2) + 10\ln(x^3-x)



    Differentiate implicitly:

    . . \frac{1}{y}\cdot y' \;=\;200\cdot\frac{2x}{x^2+2} + 10\cdot\frac{3x^2-1}{x^3-x}

    . . . . \frac{y'}{y} \;=\;10\bigg[\frac{40x}{x^2+2} + \frac{3x^2-1}{x^3-x}\bigg]

    . . . . \frac{y'}{y} \;=\;10\cdot \frac{43x^4 - 35x^2 - 2}{(x^2+2)(x^3-x)}


    . . . . y' \;=\;y\cdot10\cdot\frac{43x^4 - 35x^2 - 2}{(x^2+2)(x^3-x)}

    . . . . y'\;=\;(x^2+2)^{200}(x^3-x)^{10}\cdot 10 \cdot \frac{43x^4-35x^2-2}{(x^2+2)(x^3-x)}


    . . . . y' \;=\;10(x^2+2)^{199}(x^3-x)^9(43x^4-35x^2-2)

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  4. #4
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    thanks for your help!

    i set mine up a little differently, however

    i used (x^2 + 2)^200 as f(x)

    and (x^3 - x)^10 as g(x)

    so when i did chain rule it came out as

    f`(x) = 2x[200(x^2 + 2)^199]
    g`(x) = 3x^2[10(x^3 - x)^9]

    right so far?

    then i used product rule and came up with:

    2x[200(x^2 + 2)^199] * (x^3-x)^10 + (x^2)^200 * 3x^2[10(x^3 - x)^9]

    right? a little confused how to simplify from here. can you help?

    usually my prof lets us leave it like this but i may as well leave that as well as the way you simplified it, just in case.

    thanks again!
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