Wednesday, December 4, 2019

Calculus of a Single Variable, Chapter 2, 2.5, Section 2.5, Problem 25

(x+y)^3=x^3+y^3
Differentiating both sides with respect to x,
3(x+y)^2d/dx(x+y)=3x^2+3y^2dy/dx
3(x+y)^2(1+dy/dx)=3x^2+3y^2dy/dx
3(x+y)^2+3(x+y)^2dy/dx=3x^2+3y^2dy/dx
((x+y)^2-y^2)dy/dx=x^2-(x+y)^2
(x^2+y^2+2xy-y^2)dy/dx=(x^2-x^2-y^2-2xy)
(x^2+2xy)dy/dx=(-y^2-2xy)
dy/dx=(-y^2-2xy)/(x^2+2xy)
Derivative at point (-1,1) can be found by plugging in the values of (x,y) in dy/dx.
Therefore derivative at point (-1,1)=(-1^2-2(-1)(1))/(1^2+2(-1)(1))
Derivative = -1

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