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Math Help - composite derivative functions

  1. #1
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    Exclamation composite derivative functions

    Write the composite function in the form f(g(x)). (Identify the inner function u=g(x) and the other function y=f(u).) Then find the derivative dy/dx.

    1. y=sin(e^x)

    2. y=(1-x^2)^10

    I know the answer to number two is -20x(1-x^2)^9 but i don't know how to get to either answer.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by thecount View Post
    2. y=(1-x^2)^10

    I know the answer to number two is -20x(1-x^2)^9 but i don't know how to get to either answer.
    since you know the answer, i will do this one. i leave the other to you.

    what they are getting at is using the chain rule. it says \frac d{dx} f(g(x)) = f'(g(x)) \cdot g'(x)

    so g(x) is the function that is plugged in to f(x)

    from ({\color{red}1 - x^2})^{10}

    you should be able to see, the red is what is plugged in, in a way that makes the derivative intuitive

    so f(x) = x^{10} and g(x) = 1 - x^2

    thus, f'(x) = 10x^9 and g'(x) = -2x

    and f'(g(x)) = 10(g(x))^9 = 10(1 - x^2)^9

    so finally,

    \frac d{dx} (1 - x^2)^{10} = \underbrace{10(1 - x^2)^9}_{f'(g(x))} \cdot \underbrace{-2x}_{g'(x)} = -20x(1 - x^2)^9

    so that's the long drawn out way that they expect you to do it. you won't need to do any of this when you get used to using the chain rule. so bare this for now
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