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Math Help - Derivatives of Inverse Trig Functions

  1. #1
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    Derivatives of Inverse Trig Functions

    I'm pretty sure I'm doing this correctly, but I'm still getting a wrong answer. Any help would be appreciated.

    Find the derivative of the function. Simplify if possible.

    I'm getting:
    (\frac{1}{1+(x+\sqrt{1+x^2})^2})\times(1+\frac{1}{  2\sqrt{1+x^2}})\times2x
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by abel2 View Post
    I'm pretty sure I'm doing this correctly, but I'm still getting a wrong answer. Any help would be appreciated.

    Find the derivative of the function. Simplify if possible.

    I'm getting:
    (\frac{1}{1+(x+\sqrt{1+x^2})^2})\times(1+\frac{1}{  2\sqrt{1+x^2}})\times2x
    Here's an easy way to check this.

    if y=\text{arctan}\left(x+\sqrt{1+x^2}\right) then you'd agree that \tan(y)=x+\sqrt{1+x^2}, yes? So, differentiating both sides we get \sec^2(y)y'=1+\frac{x}{\sqrt{1+x^2}}\implies y'=\cos^2(y)\left(1+\frac{x}{\sqrt{1+x^2}}\right). Thus, y'=\cos^2\left(\arctan\left(x+\sqrt{1+x^2}\right)\  right)\left(1+\frac{x}{\sqrt{1+x^2}}\right)=\frac{  1+\frac{x}{\sqrt{1+x^2}}}{(x+\sqrt{1+x^2})^2+1}. Does that garee with your answer?
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  3. #3
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    Quote Originally Posted by abel2 View Post
    I'm pretty sure I'm doing this correctly, but I'm still getting a wrong answer. Any help would be appreciated.

    Find the derivative of the function. Simplify if possible.

    I'm getting:
    (\frac{1}{1+(x+\sqrt{1+x^2})^2})\times(1+\frac{1}{  2\sqrt{1+x^2}})\times2x
    From what I see, the 2x is in the wrong place. I get

    \left(\frac{1}{1+(x+\sqrt{1+x^2})^2}\right)\times(  1+\frac{{\color{red}{2x}}}{2\sqrt{1+x^2}})?
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