How do I find dy/dx for x2y+xy2=6 using implicit differentiation?

  • Thread starter Jim4592
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In summary, the problem is to find the derivative of x2y+xy2=6, which can be solved using the product and chain rules. The final answer is (2xy + y2)/(x2 + 2xy).
  • #1
Jim4592
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Homework Statement


find dy/dx for x2y+xy2=6


Homework Equations





The Attempt at a Solution


d/dx (x2y+xy2) = d/dx (6)
x2*(dy/dx)+y*2x+x*2y(dy/dx)+2y=0
x2*(dy/dx)+y*2x+x*2y(dy/dx)=-2y
(dy/dx)+y*2x+x*2y(dy/dx)=(-2y/x2)
(dy/dx)+y*2x+(dy/dx)=(-2y/x2+2xy)

I'm not sure what to do now, i know that the answer is (2xy-y2/x2+2xy)

I got the denominator correct in my answer but I'm not sure how to get the numerator correctly. Did I go wrong somewhere during the process?
 
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  • #2
Now you algebraically solve for dy/dx.
 
  • #3
Jim4592 said:

Homework Statement


find dy/dx for x2y+xy2=6

The Attempt at a Solution


d/dx (x2y+xy2) = d/dx (6)
x2*(dy/dx)+y*2x+x*2y(dy/dx)+2y=0
Try that again. Piece by piece:
[tex]x^2y[/tex] differentiates to [tex]2xy + x^2y'[/tex]
[tex]xy^2[/tex] differentiates to [tex]y^2 + 2xyy'[/tex]

Then follow Tom's advice!
 
  • #4
Jim4592 said:
I'm not sure what to do now, i know that the answer is (2xy-y2/x2+2xy)

I got the denominator correct in my answer but I'm not sure how to get the numerator correctly.

Hi Jim4592! :smile:

Isn't it (2xy + y2/x2+2xy)?
 
  • #5
tiny-tim said:
Isn't it (2xy + y2/x2+2xy)?
Rather,

-(2xy + y2)/(x2 + 2xy)
 
  • #6
oops! :redface:

thanks, Unco! :smile:
 

What is implicit differentiation?

Implicit differentiation is a mathematical technique used to find the derivative of a function that is not explicitly expressed in terms of a single variable. It is commonly used when the given equation cannot be easily solved for one variable.

How is implicit differentiation different from explicit differentiation?

In explicit differentiation, the function is expressed in terms of a single variable, making it easier to find the derivative. In implicit differentiation, the function is not explicitly expressed in terms of a single variable, so the chain rule is used to find the derivative.

When is implicit differentiation used?

Implicit differentiation is used when the given equation is not easily solvable for one variable, or when the function involves multiple variables. It can also be used to find the slope of a curve at a specific point.

What is the process of implicit differentiation?

The process of implicit differentiation involves taking the derivative of both sides of an equation with respect to the variable of interest. The chain rule is used to find the derivative of the function, and the derivative of the variable with respect to itself is equal to 1.

What are some common mistakes when using implicit differentiation?

Some common mistakes when using implicit differentiation include not properly applying the chain rule, forgetting to take the derivative of the variable with respect to itself, and incorrectly simplifying the resulting equation. It is important to carefully follow the steps and pay attention to the chain rule when using implicit differentiation.

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