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Math Help - d/dx f[1/g(x)] ?

  1. #1
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    d/dx f[1/g(x)] ?

    I have a question stating:

    Find d/dx f[1/g(x)]

    g(x) and f(x) are differentiable. What exactly does this mean?

    My solution is f'(1/g(x)*g'(x)/g(x)^2

    Using the chain rule there ^
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by SportfreundeKeaneKent View Post
    I have a question stating:

    Find d/dx f[1/g(x)]

    g(x) and f(x) are differentiable. What exactly does this mean?

    My solution is f'(1/g(x)*g'(x)/g(x)^2

    Using the chain rule there ^
    Yes, this is the answer

    Differentiable means that you can differentiate the function. More generally, and in this case, it means that g'(x) and f'(x) exist. Here is an example :
    f(x)=\sqrt{x}, for x \in [0, \infty[

    f'(x)=\frac{1}{2 \sqrt{x}}
    This is defined all over ]0, \infty[. But not in 0.
    The function is not differentiable in 0.

    But here you have to add that g(x) never annulates.
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  3. #3
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    I have a question stating:

    Find d/dx f[1/g(x)]

    g(x) and f(x) are differentiable. What exactly does this mean?

    My solution is f'(1/g(x)*g'(x)/g(x)^2

    Using the chain rule there ^
    Quote Originally Posted by Moo View Post
    Hello,



    Yes, this is the answer

    [snip]
    The derivative of \frac{1}{g(x)} with respect to x is {\color{red}-} \frac{g'(x)}{[g(x)]^2} so I think there might be a negative in there somewhere, too .....
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