Derivatives of Inverse Trig Functions

**abel2** · Jun 8th 2010, 04:55 PM

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:
$(\frac{1}{1+(x+\sqrt{1+x^2})^2})\times(1+\frac{1}{ 2\sqrt{1+x^2}})\times2x$

**Drexel28** · Jun 8th 2010, 05:31 PM

Originally Posted by abel2

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?

**Jester** · Jun 9th 2010, 03:34 AM

Originally Posted by abel2

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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