# Thread: A hard inverse tan partial derivative

1. ## A hard inverse tan partial derivative

a is a constant. My brain is fried from two days of aerodynamics, i cant get this one.

2. ## Inverse tan

Hello flawless
Originally Posted by flawless
a is a constant. My brain is fried from two days of aerodynamics, i cant get this one.
$z = \tan^{-1}u$

$\Rightarrow \tan z = u$

$\Rightarrow \frac{du}{dz} = \sec^2 z = 1 + \tan^2 z = 1 + u^2$

$\frac{dz}{du}= \frac{1}{1+u^2}$

For the partial derivative, assume $y$ is constant (as well as $a$) and differentiate with respect to $x$. Can you complete it now?