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What is the implicit derivative of #1= xy-(x^2-y)^2+2y #?

Answer 1

William Abdalla

#(dy/dx)=(2x-y)/(x+2y+2)#

#1=xy-(x^2-y^2)+2y# Rewritten: #1=xy-x^2+y^2+2y#

#0=x(dy/dx)+y-2x+2y(dy/dx)+2(dy/dx)# #0=(dy/dx)(x+2y+2)+y-2x# #(dy/dx)(x+2y+2)=2x-y#

#(dy/dx)=(2x-y)/(x+2y+2)#

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Answer 2

Isabella Ackerman

To find the implicit derivative of the equation 1 = xy - (x^2 - y)^2 + 2y with respect to x, we differentiate both sides of the equation with respect to x, treating y as a function of x.

Differentiating each term: d/dx(1) = d/dx(xy) - d/dx((x^2 - y)^2) + d/dx(2y)

Differentiating each term with respect to x: 0 = y + x(dy/dx) - 2(x^2 - y)(2x - dy/dx) + 2(dy/dx)

Simplify and solve for dy/dx: 0 = y + x(dy/dx) - 4x(x^2 - y) + 4y(x) - 2(dy/dx) dy/dx = (y - 4x(x^2 - y) + 4xy) / (x - 2(x^2 - y))

So, the implicit derivative of the equation 1 = xy - (x^2 - y)^2 + 2y with respect to x is:

dy/dx = (y - 4x(x^2 - y) + 4xy) / (x - 2(x^2 - y))

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.