Answer: Hello mate!
Clairaut’s Theorem says that if you have a function f(x,y) that have defined and continuous second partial derivates in (ai, bj) ∈ A
for all the elements in A, the, for all the elements on A you get:
![\frac{d^{2}f }{dxdy}(ai,bj) = \frac{d^{2}f }{dydx}(ai,bj)](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E%7B2%7Df%20%7D%7Bdxdy%7D%28ai%2Cbj%29%20%3D%20%5Cfrac%7Bd%5E%7B2%7Df%20%7D%7Bdydx%7D%28ai%2Cbj%29)
This says that is the same taking first a partial derivate with respect to x and then a partial derivate with respect to y, that taking first the partial derivate with respect to y and after that the one with respect to x.
Now our function is u(x,y) = tan (2x + 3y), and want to verify the theorem for this, so lets see the partial derivates of u. For the derivates you could use tables, for example, using that:
![\frac{d(tan(x))}{dx} = 1/cos(x)^{2} = sec(x)^{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%28tan%28x%29%29%7D%7Bdx%7D%20%3D%201%2Fcos%28x%29%5E%7B2%7D%20%3D%20sec%28x%29%5E%7B2%7D)
![\frac{du}{dx} = \frac{2}{cos^{2}(2x + 3y)} = 2sec(2x + 3y)^{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bdu%7D%7Bdx%7D%20%20%3D%20%20%5Cfrac%7B2%7D%7Bcos%5E%7B2%7D%282x%20%2B%203y%29%7D%20%3D%202sec%282x%20%2B%203y%29%5E%7B2%7D)
and now lets derivate this with respect to y.
using that ![\frac{d(sec(x))}{dx}= sec(x)*tan(x)](https://tex.z-dn.net/?f=%20%5Cfrac%7Bd%28sec%28x%29%29%7D%7Bdx%7D%3D%20sec%28x%29%2Atan%28x%29)
![\frac{du}{dxdy} = \frac{d(2*sec(2x + 3y)^{2} )}{dy} = 2*2sec(2x + 3y)*sec(2x + 3y)*tan(2x + 3y)*3 = 12sec(2x + 3y)^{2}tan(2x + 3y)](https://tex.z-dn.net/?f=%5Cfrac%7Bdu%7D%7Bdxdy%7D%20%3D%20%5Cfrac%7Bd%282%2Asec%282x%20%2B%203y%29%5E%7B2%7D%20%29%7D%7Bdy%7D%20%20%3D%202%2A2sec%282x%20%2B%203y%29%2Asec%282x%20%2B%203y%29%2Atan%282x%20%2B%203y%29%2A3%20%3D%2012sec%282x%20%2B%203y%29%5E%7B2%7Dtan%282x%20%2B%203y%29)
Now if we first derivate by y, we get:
![\frac{du}{dy} = \frac{3}{cos^{2}(2x + 3y)} = 3sec(2x + 3y)^{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bdu%7D%7Bdy%7D%20%20%3D%20%20%5Cfrac%7B3%7D%7Bcos%5E%7B2%7D%282x%20%2B%203y%29%7D%20%3D%203sec%282x%20%2B%203y%29%5E%7B2%7D)
and now we derivate by x:
![\frac{du}{dydx} = \frac{d(3*sec(2x + 3y)^{2} )}{dy} = 3*2sec(2x + 3y)*sec(2x + 3y)*tan(2x + 3y)*2 = 12sec(2x + 3y)^{2}tan(2x + 3y)](https://tex.z-dn.net/?f=%5Cfrac%7Bdu%7D%7Bdydx%7D%20%3D%20%5Cfrac%7Bd%283%2Asec%282x%20%2B%203y%29%5E%7B2%7D%20%29%7D%7Bdy%7D%20%20%3D%203%2A2sec%282x%20%2B%203y%29%2Asec%282x%20%2B%203y%29%2Atan%282x%20%2B%203y%29%2A2%20%3D%2012sec%282x%20%2B%203y%29%5E%7B2%7Dtan%282x%20%2B%203y%29)
the mixed partial derivates are equal :)