# Thread: Derivatives of Inverse Trig Functions

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

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

if $\displaystyle y=\text{arctan}\left(x+\sqrt{1+x^2}\right)$ then you'd agree that $\displaystyle \tan(y)=x+\sqrt{1+x^2}$, yes? So, differentiating both sides we get $\displaystyle \sec^2(y)y'=1+\frac{x}{\sqrt{1+x^2}}\implies y'=\cos^2(y)\left(1+\frac{x}{\sqrt{1+x^2}}\right)$. Thus, $\displaystyle y'=\cos^2\left(\arctan\left(x+\sqrt{1+x^2}\right)\ right)\left(1+\frac{x}{\sqrt{1+x^2}}\right)=\frac{ 1+\frac{x}{\sqrt{1+x^2}}}{(x+\sqrt{1+x^2})^2+1}$. Does that garee with your answer?

3. 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:
$\displaystyle (\frac{1}{1+(x+\sqrt{1+x^2})^2})\times(1+\frac{1}{ 2\sqrt{1+x^2}})\times2x$
From what I see, the $\displaystyle 2x$ is in the wrong place. I get

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