Find dx/dy by implicit differentiation
x^2y+y^2x
P.S: It's calculus
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)]
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)
y = x^2y+y^2x
dy/dx = 2y X^2y-1 + 2x Y^2x-1
To find dx/dy by implicit differentiation, you'll need to differentiate both sides of the equation with respect to y, treating x as a function of y. Let's assume the equation is:
x^2y + y^2x = 0
To differentiate this equation implicitly, follow these steps:
Step 1: Differentiate both terms with respect to y using the product rule and chain rule.
For the term x^2y, apply the product rule:
d/dy (x^2y) = 2xy + x^2(dy/dx)
For the term y^2x, apply the product rule:
d/dy (y^2x) = 2yx + y^2(dx/dy)
Step 2: Combine the differentiated terms on one side of the equation.
2xy + x^2(dy/dx) + 2yx + y^2(dx/dy) = 0
Step 3: Rearrange the equation and isolate dx/dy.
x^2(dy/dx) + y^2(dx/dy) = -2xy - 2yx
Step 4: Solve for dx/dy.
To solve for dx/dy, divide both sides of the equation by y^2:
(dx/dy) = (-2xy - 2yx) / (x^2 + y^2)
So, the derivative dx/dy is given by (-2xy - 2yx) / (x^2 + y^2).
Note: Be careful to differentiate correctly using the chain rule when differentiating terms with x's. Make sure to take into account that y is treated as a function of x.