Asked by amy
implicit differentation
1.y^3=4(x^2+y^2)
2.y^2-3x+2y=0
help please
You consider y to be a function of x, but you don't explicitely solve for it. Then you formally differentiate w.r.t. x using the chain rule:
y^3(x) = 4(x^2+y^2(x)) --->
3y^2dy/dx = 8x + 8y dy/dx
y^2-3x+2y=0 --->
2ydy/dx - 3+2dy/dx = 0
1.y^3=4(x^2+y^2)
2.y^2-3x+2y=0
help please
You consider y to be a function of x, but you don't explicitely solve for it. Then you formally differentiate w.r.t. x using the chain rule:
y^3(x) = 4(x^2+y^2(x)) --->
3y^2dy/dx = 8x + 8y dy/dx
y^2-3x+2y=0 --->
2ydy/dx - 3+2dy/dx = 0
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