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:
[tex]\frac{d^{2}f }{dxdy}(ai,bj) = \frac{d^{2}f }{dydx}(ai,bj)[/tex]
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:
[tex]\frac{d(tan(x))}{dx} = 1/cos(x)^{2} = sec(x)^{2}[/tex]
[tex]\frac{du}{dx} = \frac{2}{cos^{2}(2x + 3y)} = 2sec(2x + 3y)^{2}[/tex]
and now lets derivate this with respect to y.
using that [tex] \frac{d(sec(x))}{dx}= sec(x)*tan(x)[/tex]
[tex]\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)[/tex]
Now if we first derivate by y, we get:
[tex]\frac{du}{dy} = \frac{3}{cos^{2}(2x + 3y)} = 3sec(2x + 3y)^{2}[/tex]
and now we derivate by x:
[tex]\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)[/tex]
the mixed partial derivates are equal :)
