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Math Help - Find the derivative of the function below.

  1. #1
    Junior Member
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    Find the derivative of the function below.

    h(y)=(b/(a+y^4))^4
    find h'(y)

    I was using the chain rule

    h(x)=f(g(x))
    h'(x)=f'(g(x))*g'(x)

    and I got...

    4(b/(a+y^4))((-4by^3)/(a+y^4)^2)

    could someone tell me the answer, or if you have time, tell me step by step what to do?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Latszer View Post
    h(y)=(b/(a+y^4))^4
    find h'(y)

    I was using the chain rule

    h(x)=f(g(x))
    h'(x)=f'(g(x))*g'(x)

    and I got...

    4(b/(a+y^4))((-4by^3)/(a+y^4)^2)

    could someone tell me the answer, or if you have time, tell me step by step what to do?
    You're so close!!

    It should be 4\left(\frac{b}{a+y^4}\right)^{{\color{red}3}}\lef  t(-\frac{4by^3}{\left(a+y^4\right)^2}\right)
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  3. #3
    Senior Member DeMath's Avatar
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    Moscow
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    Quote Originally Posted by Latszer View Post
    h(y)=(b/(a+y^4))^4
    find h'(y)

    I was using the chain rule

    h(x)=f(g(x))
    h'(x)=f'(g(x))*g'(x)

    and I got...

    4(b/(a+y^4))((-4by^3)/(a+y^4)^2)

    could someone tell me the answer, or if you have time, tell me step by step what to do?
    h'\left( y \right) =  - \frac{{4{b^4}}}{{{{\left( {a + {y^4}} \right)}^5}}} \cdot {\left( {a + {y^4}} \right)^\prime } =  - \frac{{4{b^4}}}{{{{\left( {a + {y^4}} \right)}^5}}} \cdot 4{y^3} =  - \frac{{16{b^4}{y^3}}}{{{{\left( {a + {y^4}} \right)}^5}}}.
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  4. #4
    Junior Member
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    Yeah, made an error and didn't put it to the third, I guess I had it right. Thanks you two.
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