`(x + y)^3 = x^3 + y^3, (-1,1)` Find `dy/dx` by implicit differentiation and evaluate the derivative at the given point.

Textbook Question

Chapter 2, 2.5 - Problem 25 - Calculus of a Single Variable (10th Edition, Ron Larson).
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gsarora17's profile pic

gsarora17 | (Level 2) Associate Educator

Posted on

`(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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kandukurimaths | Student, Graduate | (Level 1) Honors

Posted on

given function is

(x + y)3 = x3 + y3

using polynomial identity

(x + y)3 = x3 + 3x2y + 3xy2 + y3

we get

3x2y + 3xy2 =0

3x2y  =- 3xy2

x=-y

y=-x

dy/dx=-1 at any point

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