Show ( d(u,v)/d(x,y))(d(x,y)/d(r,s))= d(u,v)/d(r,s)
( d(u,v)/d(x,y)) = 1/ (d(x,y)/d(u,v))
hint: |AB| = |A||B|
I have dealt with partial derivative problems a lot before, but theory always throws me off. I know it would seem to solve itself just by looking at it but lol no.
- kbLv 710 years agoFavorite Answer
Note that |∂(u,v)/∂(x,y)| represents the determinant of the Jacobian Matrix,
which yields the derivative of f(x, y) = (u(x,y), v(x,y)). I'll call this matrix Df
Letting f(x, y) = (u(x,y), v(x,y)), and g(r, s) = (x(r, s), y(r, s)).
Then, the Chain Rule for Jacobian (Derivative) Matrices yields
D(f o g) = Df * Dg, where the * on the right side is matrix multiplication.
This may be alternately written as
∂(u,v)/∂(x,y) * ∂(x,y)/∂(r,s) = ∂(u,v)/∂(r,s).
Taking determinants of both sides:
|∂(u,v)/∂(x,y) * ∂(x,y)/∂(r,s)| = |∂(u,v)/∂(r,s)|
==> |∂(u,v)/∂(x,y)| * |∂(x,y)/∂(r,s)| = |∂(u,v)/∂(r,s)|, using the hint.
The second result follows from letting u = r and v = s in the first result.
==> |∂(u,v)/∂(x,y)| * |∂(x,y)/∂(u, v)| = |∂(u,v)/∂(u,v)|.
However, |∂(u,v)/∂(u,v)| =
|0 1| = 1.
|∂(u,v)/∂(x,y)| = 1/ |∂(x,y)/∂(u, v)|, as long as these expressions are non-zero.
I hope this helps!
- Anonymous4 years ago
deriviative is 4x-(5xy'+5y+xy) use product rule for 5xy, use x's rather derivative and write derivative of y as y' and circulate away it on my own after that. in case you didnt comprehend product rule is derivitive of one term cases the different plus the derivitive of the different term cases the 1st term thats troublesome to comprehend so der. of xy is x'y+xy' that's often written as y+xy' in terms of x considering the fact that derivative of x is a million you additionally can use the chain rule after pulling an x out so as that this is x(2x-5y) wich then you could use the product rule besides and get (2x-5y)+x(2+5y') that's 4x -5y+2x+5y' which with a touch extra artwork you could practice equals the derivative of the 1st occasion i confirmed you.