Implicit differentiation

  • Thread starter callmelish
  • Start date
  • #1

Homework Statement


"Find dy/dx at the given point by using implicit differentiation"

x2y + y2x = -2 at (2, -1)

and

(x+y)3 = x3 + y3


Homework Equations





The Attempt at a Solution


1) x2(dy/dx) + y(2x) + y2(1) + 2y(dy/dx)(x) = -2
x2(dy/dx) + 2xy + y2 + 2xy(dy/dx) = -2
dy/dx(x2 + 2xy) = 2xy + y2 -2
dy/dx = (2xy + y2 -2)/(x2 + 2xy)

dy/dx at (2, -1) = (2*2*-1 - 2)/(22 + 2*2*-1)
= 0/0 = 0

The first one has me confused, and since the second one is similar I didn't want to attempt it in case I'm completely wrong.
 

Answers and Replies

  • #2
264
0
"x^2y + y^2x = -2 at (2, -1)"

When you take the derivative of this, whether implicit or not, -2 will fall out of the equation (go to zero).
 
  • #3
dextercioby
Science Advisor
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Insights Author
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Don't know that

[tex] \frac{dy}{dx} = -\frac{\frac{\partial F(x,y)}{\partial x}}{\frac{\partial F(x,y)}{\partial y}} [/tex]

, where F(x,y)=0 is the implicit equation ?
 
  • #4
eumyang
Homework Helper
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1) x2(dy/dx) + y(2x) + y2(1) + 2y(dy/dx)(x) = -2
The bolded should be 0 (the derivative of -2 is 0).
x2(dy/dx) + 2xy + y2 + 2xy(dy/dx) = -2
dy/dx(x2 + 2xy) = 2xy + y2 -2
Both of the bolded terms here should be negative.
 
  • #5
209
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What is the derivative of the constant function f(x) = -2? Don't forget you have to differentiate both sides of the equation.
 
  • #6
264
0
Also it may help to look at an implicit plot of this function, at (2,-1) the slope is asymptotic.
 

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