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Math Help - product rule, taking derivatives

  1. #1
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    product rule, taking derivatives

    2. Find the derivative of   y = 3 sin^{3} (2x^{4} + 1)

    I used the product rule and this is my working, however I don't end up with the correct answer, can someone please tell me where I am going wrong, thanks.

     u = 3sin^{3}

     v = 2x^{4} +1

     t = sin , y= 3t^{3}


     \frac{dt}{dx} = cos

     \frac{dy}{dt} = 9t^{2}

    so now I multiply both together to get the derivative of  3sin^{3}

     9sin^{2} \times cos

     \frac{dy}{dx} = 8x^{3}

    Now using the product rule and putting this together.


     9sin^{2} \times cos \times (2x^{4} +1) + 8x^{3} \times 3sin^{3}

    the correct answer is.
     72x^{3} sin^{2}(2x^{4} +1)cos(2x^{4} +1)
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  2. #2
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    not the product rule ... the chain rule

    y = 3[\sin(2x^4+1)]^3

     <br />
y' = 9[\sin(2x^4+1)]^2 \cdot \cos(2x^4+1) \cdot 8x^3<br />

    now clean up the algebra
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  3. #3
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    Quote Originally Posted by skeeter View Post
    not the product rule ... the chain rule

    y = 3[\sin(2x^4+1)]^3

     <br />
y' = 9[\sin(2x^4+1)]^2 \cdot \cos(2x^4+1) \cdot 8x^3<br />

    now clean up the algebra
    Oh I see.

    But I am just wondering how did you know to use the chain rule instead of the product rule? As this expression is a product of two functions, so that's why I assumed I should use the product rule.

    and I thought the chain rules was supposed to be used for a function of a function. Like  \sqrt{3x-1}

    So for an expression like this  x^{2} \sqrt{3x-1}

    would you use the chain rule or product rule?

    thank you for clarifying, greatly appreciated.
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  4. #4
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    Quote Originally Posted by Tweety View Post
    Oh I see.

    But I am just wondering how did you know to use the chain rule instead of the product rule? As this expression is a product of two functions, <<<< No. See below
    so that's why I assumed I should use the product rule.

    and I thought the chain rules was supposed to be used for a function of a function. <<<<< Correct

    ...
    Your function is indeed the function of a function of a function:

    f_1(x)=2x^4+1

    f_2(x)=\sin(x)

    f_3(x)=3x^3

    And your function becomes:

    f(x)=f_3(f_2(f_1(x)))

    As Skeeter indicated the powers of trigonometric functions are written at the name of the function:

    \sin^2(x)=(\sin(x))^2

    and - very important - mostly this unequality is true:

    \sin^2(x) \neq \sin((x)^2)
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