Verify that the conclusion of Clairaut’s Theorem holds, that is, uxy = uyx, u=tan(2x+3y)

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) ∈ Afor 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 :)