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

Answer 1

Elijah Abram

#(dy)/(dx)=-(2xy+y^2}/(x^2+2xy)# and at #(-1,1)#, #(dy)/(dx)=-1#

As #(x+y)^3=x^3+y^3#, taking differential on both sides implicitly, we get

#3(x+y)^2xx(1+(dy)/(dx))=3x^2+3y^2(dy)/(dx)#

or #3(dy)/(dx)=3{x^2-(x+y)^2}#

or #(dy)/(dx)={x^2-(x+y)^2}/{(x+y)^2-y^2}#

or #(dy)/(dx)=-(2xy+y^2}/(x^2+2xy)#

and at #(-1,1)#

#(dy)/(dx)=-(2(-1)xx1+1^2}/(1^2+2(-1)(1))=-(-1)/(-1)=-1#

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Answer 2

Cameron Alexander

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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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.