# How do you find #dy/dx# by implicit differentiation of #(x+y)^3=x^3+y^3# and evaluate at point (-1,1)?

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To find dy/dx by implicit differentiation of the equation (x+y)^3=x^3+y^3 and evaluate it at the point (-1,1), you would differentiate both sides of the equation with respect to x, then solve for dy/dx, and finally substitute x = -1 and y = 1 into the resulting expression. The derivative is dy/dx = (3x^2 - 3y^2) / (3y^2 - 3x^2), and when evaluated at (-1,1), dy/dx = -4/3.

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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.

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.

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.

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