Asked by Skye
Find dx/dy by implicit differentiation
x^2y+y^2x
P.S: It's calculus
x^2y+y^2x
P.S: It's calculus
Answers
Answered by
drwls
Differentiate with respect to y, recognizing that x is a function of y.
You cannot "find" dx/dy unless you write an equation. You have only written a formula for a function g(x,y).
If x^2*y + y^2*x = C (any constant), then
x^2 + y*2x*dx/dy + 2yx + y^2 dx/dy = 0
dx/dy(y^2x + y^2) = -x^2 -2yx
dx/dy = -x(x-2y)/[y^2(1+x)]
You cannot "find" dx/dy unless you write an equation. You have only written a formula for a function g(x,y).
If x^2*y + y^2*x = C (any constant), then
x^2 + y*2x*dx/dy + 2yx + y^2 dx/dy = 0
dx/dy(y^2x + y^2) = -x^2 -2yx
dx/dy = -x(x-2y)/[y^2(1+x)]
Answered by
Damon
I think you want this equal to a constant if you want a solutionyou want dx/dy so differentiate wrt y
x^2 + y 2 x dx/dy+ y^2 dx/dy + x 2 y
= x^2 + 2 xy dx/dy + y^2 dx/dy + 2 x y
= dx/dy (2xy + y^2) + x^2 + 2xy
now if the right hand were a constant, then that = 0 so
dx/dy = -(x^2+2xy)/(y^2+2xy)
x^2 + y 2 x dx/dy+ y^2 dx/dy + x 2 y
= x^2 + 2 xy dx/dy + y^2 dx/dy + 2 x y
= dx/dy (2xy + y^2) + x^2 + 2xy
now if the right hand were a constant, then that = 0 so
dx/dy = -(x^2+2xy)/(y^2+2xy)
Answered by
Anonymous
y = x^2y+y^2x
dy/dx = 2y X^2y-1 + 2x Y^2x-1
dy/dx = 2y X^2y-1 + 2x Y^2x-1
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