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Math Help - Chain rule using trig functions

  1. #1
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    Chain rule using trig functions

    Find the derivative: (do not simplify)
    f(x) = sec(sqrt(1+(2x/tanx)))

    Is this the correct answer?
    tan (sqrt(1+(2x/tanx))) sec (sqrt(1+(2x/tanx)))1/2(1+(2x/tanx)))^-1/2 ((2tanx-2xsec^x)/(tan^2x))

    sorry I don't know how to make this look any better.
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  2. #2
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    Re: Chain rule using trig functions

    Quote Originally Posted by tmas View Post
    Find the derivative: (do not simplify)
    f(x) = sec(sqrt(1+(2x/tanx)))
    Is this the correct answer?
    tan (sqrt(1+(2x/tanx))) sec (sqrt(1+(2x/tanx)))1/2(1+(2x/tanx)))^-1/2 ((2tanx-2xsec^x)/(tan^2x))

    sorry I don't know how to make this look any better.
    Almost, The last factor is \frac{2\tan(x)-2x\sec^2(x)}{\tan^2(x)}
    Last edited by Plato; May 27th 2012 at 10:41 AM.
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  3. #3
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    Re: Chain rule using trig functions

    Why is it x^2 in the second part of the subtraction.. before taking the derivative isn't the numerator:

    (2x)'(tanx) - 2x(tanx)' ? I do not quite understand...
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  4. #4
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    Re: Chain rule using trig functions

    That was a typo.
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  5. #5
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    Re: Chain rule using trig functions

    Hello, tmas!

    \text{Find the derivative (do not simplify): }\:f(x) \:=\: \sec\!\left(\!\sqrt{1+\frac{2x}{\tan x}}\right)

    \text{Is this the correct answer?}

    . f'(x) \;=\;\tan\!\left(\!\sqrt{1+\frac{2x}{\tan x}}}\right)\sec\!\left(\!\sqrt{1+\frac{2x}{\tan x}}\right) \cdot\frac{1}{2}\cdot \left(1+\frac{2x}{\tan x}\right)^{-\frac{1}{2}}\!\left(\frac{2\tan x-2x\sec^2x}{\tan^2x}\right)

    Correct! . . . Good work!
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