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Math Help - Differentiating Inverse Functions

  1. #1
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    Differentiating Inverse Functions

    Hey guys I'm having problems with this question, could you guys help me out?

    The question is:
    Show that \dfrac{d}{dx}\left(\tan^{-1}(\frac{3}{2}\tan x)\right)=\dfrac{6}{5\sin^2 x+4}

    heres what I did
    \dfrac{d}{dx}\left(\tan^{-1}(\frac{3}{2}\tan x)\right)

    =\dfrac{\dfrac{3}{2}\times\dfrac{1}{1+x^2}}{1+\lef  t(\dfrac{3}{2}\tan x\right)^2}

    =\dfrac{\dfrac{3}{2(1+x^2)}}{1+\dfrac{9sin^2x}{4co  s^2x}}

    <br />
=\dfrac{\dfrac{3}{2(1+x^2)}}{\dfrac{4cos^2x+9sin^2  x}{4cos^2x}}


    =\dfrac{6cos^2x}{(1+x^2)(4cos^2x+9sin^2x)}

    <br />
=\dfrac{6cos^2x}{(1+x^2)(5sin^2x+4)}

    this kinda seems quite close to the result that I had to prove, did I do something wrong?

    Thanks
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by aonin View Post
    =\dfrac{\dfrac{3}{2}\times\dfrac{1}{1+x^2}}{1+\lef  t(\dfrac{3}{2}\tan x\right)^2}

    It should be:

    \dfrac{\dfrac{3}{2}\dfrac{1}{\cos^2 x}}{1+\left(\dfrac{3}{2}\tan x\right)^2}
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  3. #3
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    If \displaystyle y = \arctan{\left(\frac{3}{2}\tan{x}\right)}

    then \displaystyle \tan{y} = \frac{3}{2}\tan{x}

    \displaystyle \frac{d}{dx}(\tan{y}) = \frac{d}{dx}\left(\frac{3}{2}\tan{x}\right)

    \displaystyle \frac{d}{dy}(\tan{y})\,\frac{dy}{dx} = \frac{3}{2}\sec^2{x}

    \displaystyle \sec^2{y}\,\frac{dy}{dx} = \frac{3}{2}\sec^2{x}

    \displaystyle (1 + \tan^2{y})\,\frac{dy}{dx} = \frac{3}{2}\sec^2{x}

    \displaystyle \left\{1 + \tan^2{\left[\arctan{\left(\frac{3}{2}\tan{x}\right)}\right]\right\}\,\frac{dy}{dx} = \frac{3}{2}\sec^2{x}

    \displaystyle \left[1 + \left(\frac{3}{2}\tan{x}\right)^2\right]\,\frac{dy}{dx} = \frac{3}{2}\sec^2{x}

    \displaystyle \frac{dy}{dx} = \frac{\frac{3}{2}\sec^2{x}}{1 + \left(\frac{3}{2}\tan{x}\right)^2}
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  4. #4
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    Wink

    Quote Originally Posted by FernandoRevilla View Post
    It should be:

    \dfrac{\dfrac{3}{2}\dfrac{1}{\cos^2 x}}{1+\left(\dfrac{3}{2}\tan x\right)^2}
    do you mind explaining this step, I didn't really get it

    when doing the question I used this formula
    \dfrac{d}{dx}tan^{-1}f(x)=\dfrac{f'(x)}{1+[f(x)]^{2}}

    so wouldn't f'(x) be u'v+v'u=0\times tanx+\dfrac{1}{1+x^2}\times \dfrac{3}{2}=\dfrac{3}{2(1+x^2)}

    or am I misunderstanding something?

    Thanks
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by aonin View Post
    when doing the question I used this formula
    \dfrac{d}{dx}tan^{-1}f(x)=\dfrac{f'(x)}{1+[f(x)]^{2}}

    All right, and f(x)=(3/2) \tan x so, f'(x)=(3/2)(\tan x)'=(3/2)\sec ^2 x .
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  6. #6
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    Quote Originally Posted by aonin View Post
    do you mind explaining this step, I didn't really get it

    when doing the question I used this formula
    \dfrac{d}{dx}tan^{-1}f(x)=\dfrac{f'(x)}{1+[f(x)]^{2}}

    so wouldn't f'(x) be u'v+v'u=0\times tanx+\dfrac{1}{1+x^2}\times \dfrac{3}{2}=\dfrac{3}{2(1+x^2)}
    You differentiated tan^{-1}(x) again, not tan(x).

    or am I misunderstanding something?

    Thanks
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  7. #7
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    alright I get it now, careless mistake

    Thanks a lot
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  8. #8
    Ted
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    Keep this in your mind:

    \displaystyle \frac{d}{dx} \left( arctan(f(x)) \right) = \frac{f'(x)}{1+\left[ f(x) \right]^2}
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