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Math Help - did i do this correctly?

  1. #1
    Senior Member
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    did i do this correctly?

    find f'(\frac{\pi}{4})
    if f(x)=\frac{2\sqrt{x}}{tan x}

    my answer: f'(\frac{\pi}{4})=\frac{2-4\pi}{\sqrt{\pi}}
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by yoman360 View Post
    find f'(\frac{\pi}{4})
    if f(x)=\frac{2\sqrt{x}}{tan x}

    my answer: f'(\frac{\pi}{4})=\frac{2-4\pi}{\sqrt{\pi}}
    Close. The derivative is of f is:

    \frac{\tan x\cdot\frac{1}{\sqrt{x}}-2\sqrt{x}\cdot\sec^2x}{\tan^2x}

    Plugging in \frac{\pi}{4} gives us

     \frac{\tan(\pi/4)\cdot\frac{1}{\sqrt{\pi/4}}-2\sqrt{\pi/4}\cdot\sec^2(\pi/4)}{\tan^2(\pi/4)}=\frac{\frac{2}{\sqrt{\pi}}-\sqrt{\pi}\cdot2}{1} = \boxed{\frac{2-2\pi}{\sqrt{\pi}}}
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  3. #3
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    Quote Originally Posted by redsoxfan325 View Post
    Close. The derivative is of f is:

    \frac{\tan x\cdot\frac{1}{\sqrt{x}}-2\sqrt{x}\cdot\sec^2x}{\tan^2x}

    Plugging in \frac{\pi}{4} gives us

     \frac{\tan(\pi/4)\cdot\frac{1}{\sqrt{\pi/4}}-2\sqrt{\pi/4}\cdot\sec^2(\pi/4)}{\tan^2(\pi/4)}=\frac{\frac{2}{\sqrt{\pi}}-\sqrt{\pi}\cdot2}{1} = \boxed{\frac{2-2\pi}{\sqrt{\pi}}}
    Ok thanks 1 little error messed me up. Thanks
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