# Derivative of Inverse Trigonometric Function

• Mar 28th 2009, 04:56 AM
Derivative of Inverse Trigonometric Function
http://www.webassign.net/www22/symIm...72a34582c6.gif

Find the derivative...

I know d/dx(arctan u)= u'/1+u^2

Can anyone help with this?
• Mar 28th 2009, 06:27 AM
Scott H
You are correct that $\frac{d}{dx}\arctan u=\frac{u'}{1+u^2}$. Applying this formula to $y$, we obtain

$\frac{dy}{dx}=\frac{d}{dx}\left(\arctan\left(\frac {x}{5}\right)-\frac{1}{2(x^2+25)}\right)$
$=\frac{d}{dx}\arctan\left(\frac{x}{5}\right)-\frac{d}{dx}\left(\frac{1}{2(x^2+25)}\right)$
$=\frac{\frac{d}{dx}\left(\frac{x}{5}\right)}{1+\le ft(\frac{x}{5}\right)^2}+\frac{1}{(2(x^2+25))^2}\c dot(4x)$
$=\frac{\frac{1}{5}}{1+\frac{x^2}{25}}+\frac{x}{(x^ 2+25)^2}$
$=\frac{5}{25+x^2}+\frac{x}{(x^2+25)^2}$
$=\frac{5(x^2+25)}{(x^2+25)^2}+\frac{x}{(x^2+25)^2}$
$=\frac{5x^2+125}{(x^2+25)^2}+\frac{x}{(x^2+25)^2}$
$=\frac{5x^2+x+125}{(x^2+25)^2}.$