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Thread: Suppose that f and g are differentiable functions with the values given below...?

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    Suppose that f and g are differentiable functions with the values given below...?

    Suppose that f and g are differentiable functions with the values given below. If h(x)= f {g[sqrt(x)]} and k(x)= g[ 1/x ・f(2x)], show that h'(4)=3 and k'(1)=16..

    x=2, f(x)=2, g(x)=2, f'(x)=3, g'(x)=4

    I have no idea with this question.. Can anyone help me?? Please!!!
    Last edited by Yukina; Jul 6th 2011 at 06:05 PM.
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  2. #2
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    Re: Suppose that f and g are differentiable functions with the values given below...?

    what is your question and i don't understand clearly on k(x)= g[ 1/xf(2x)],
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    Re: Suppose that f and g are differentiable functions with the values given below...?

    Quote Originally Posted by Yukina View Post
    Suppose that f and g are differentiable functions with the values given below. If h(x)= f {g[sqrt(x)]} and k(x)= g[ 1/xf(2x)], show that h'(4)=3 and k'(1)=16..

    x=2, f(x)=2, g(x)=2, f'(x)=3, g'(x)=4

    I have no idea with this question
    heard of the chain rule?

    $\displaystyle h(x) = f[g(\sqrt{x})]$

    $\displaystyle h'(x) = f'[g(\sqrt{x})] \cdot g'(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}$

    evaluate $\displaystyle h'(4)$


    please clarify the expression for $\displaystyle k(x)$ ... what you posted is a mess.
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    Re: Suppose that f and g are differentiable functions with the values given below...?

    Hello, Yukina!

    Suppose that $\displaystyle f(x)$ and $\displaystyle g(x)$ are differentiable functions

    with these values: .$\displaystyle \begin{Bmatrix}f(2) &=& 2 \\ g(2) &=& 2 \\ f'(2) &=&3 \\ g'(2) &=& 4 \end{Bmatrix}$

    $\displaystyle \text{If }h(x)\:=\:f[g(\sqrt{x})]\,\text{ and }\,k(x) \:=\: g\left[\tfrac{1}{x}\!\cdot\!f(2x)],$

    . . $\displaystyle \text{ show that: }\: h'(4)\:=\:3\,\text{ and }\,k'(1)\:=\:16$

    $\displaystyle \text{Find }h'(x)\!:\;h'(x) \;=\;f'[g(\sqrt{x})]\cdot g'(\sqrt{x}) \cdot \tfrac{1}{2}x^{-\frac{1}{2}} $

    $\displaystyle \text{Then: }\:h'(4) \;=\;f'[g(\sqrt{4})]\cdot g'(\sqrt{4})\cdot \frac{1}{2}\cdot\frac{1}{\sqrt{4}} $

    . . . . . . . . .$\displaystyle =\;f'(g(2)] \cdot g'(2) \cdot\frac{1}{2}\cdot\frac{1}{2} $

    . . . . . . . . .$\displaystyle =\;f'(2) \cdot 4\cdot \frac{1}{4}$

    . . . . . . . . .$\displaystyle =\;3\cdot 4\cdot \frac{1}{4}$

    . . . . . . . . .$\displaystyle =\;3$



    $\displaystyle \text{Find }k'(x)\!:\;k'(x) \;=\;g'\left[\tfrac{1}{x}\!\cdot\!f(2x)\right]\!\cdot\!\left[\tfrac{1}{x}\!\cdot\!f'(2x)\!\cdot\!2 - \tfrac{1}{x^2}\!\cdot\!f(2x)\right] $

    $\displaystyle \text{Then: }\:k'(1) \;=\;g'[\tfrac{1}{1}\!\cdot\!f(2)]\!\cdot\!\left[\tfrac{1}{1}\!\cdot\!2\!\cdot\! f'(2) - \tfrac{1}{1}\!\cdot\!f(2)\right] $

    . . . . . . . . . $\displaystyle =\;g'(2)\left[2\!\cdot\!f'(2) - f(2)\right]$

    . . . . . . . . . $\displaystyle =\;4\!\cdot\!\left[2\!\cdot\!3 - 2\right]$

    . . . . . . . . . $\displaystyle =\;4\!\cdot\!4$

    . . . . . . . . . $\displaystyle =\;16$

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