What is the Implicit Derivative at a Given Point?

In summary, the problem is to find the derivative of the implicit equation x^2y + y^2x = -2 at the point (2, -1), using implicit differentiation. The solution involves taking the derivative of both sides of the equation and using the chain rule, which results in dy/dx = (2xy + y^2 - 2)/(x^2 + 2xy). However, at the given point, the slope is undefined due to a vertical tangent, resulting in a division by zero.
  • #1
callmelish
1
0

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.
 
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  • #2
"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
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
callmelish said:
1) x2(dy/dx) + y(2x) + y2(1) + 2y(dy/dx)(x) = -2
The bolded should be 0 (the derivative of -2 is 0).
callmelish said:
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
What is the derivative of the constant function f(x) = -2? Don't forget you have to differentiate both sides of the equation.
 
  • #6
Also it may help to look at an implicit plot of this function, at (2,-1) the slope is asymptotic.
 

Related to What is the Implicit Derivative at a Given Point?

1. What is implicit differentiation?

Implicit differentiation is a mathematical technique used to find the derivative of a function that is not explicitly written in terms of one variable. It is particularly useful for finding the derivative of equations that cannot be easily solved for one variable.

2. How is implicit differentiation different from explicit differentiation?

Explicit differentiation involves finding the derivative of a function that is written explicitly in terms of one variable. Implicit differentiation, on the other hand, is used when the function is not explicitly written in terms of one variable and involves differentiating both sides of an equation with respect to a variable.

3. When is implicit differentiation used?

Implicit differentiation is used when a function is difficult or impossible to solve explicitly for one variable. It is also used when a function is given as an equation rather than a function in terms of a variable.

4. What are the steps for implicit differentiation?

The steps for implicit differentiation are:

  • Differentiate both sides of the equation with respect to the variable you are solving for.
  • Use the chain rule for any terms that involve that variable.
  • Solve for the derivative of the variable in terms of other variables.

5. What are some common mistakes made in implicit differentiation?

Some common mistakes made in implicit differentiation include:

  • Forgetting to use the chain rule for terms involving the variable being differentiated.
  • Not differentiating correctly on both sides of the equation.
  • Incorrectly solving for the derivative of the variable.

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