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

User Ebikeneser
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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:


(d^(2)f )/(dxdy)(ai,bj) = (d^(2)f )/(dydx)(ai,bj)

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:


(d(tan(x)))/(dx) = 1/cos(x)^(2) = sec(x)^(2)


(du)/(dx)  =  (2)/(cos^(2)(2x + 3y)) = 2sec(2x + 3y)^(2)

and now lets derivate this with respect to y.

using that
(d(sec(x)))/(dx)= sec(x)*tan(x)


(du)/(dxdy) = (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)

Now if we first derivate by y, we get:


(du)/(dy)  =  (3)/(cos^(2)(2x + 3y)) = 3sec(2x + 3y)^(2)

and now we derivate by x:


(du)/(dydx) = (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)

the mixed partial derivates are equal :)

User Ruprit
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