Question Involving implicit Differentiation

In summary, the person is trying to find the derivative of a fraction with a variable on the left hand side. They are confused about the numerator and denominator and are looking for help.
  • #1
skies222
6
0
Hi!
I have a problem here that's been bugging me. I was wondering if anyone can give insight into where I'm going wrong

implicit differentiation problem

1) (x^2+y^2)/(x+y)=xy-2

find derivitive (dy/dx) at point (-1, -1)


I know the basic premise. I used the quotient rule to find the derivative of the fraction. I am confused about the resulting fraction though. Once you find the derivative of the fraction, what should you do about the denominator ( which will be (x+y)^2 ). do you multiply each side by it? ( which would remove it from the left hand side, but add it to the right hand side). I've also tried moving the "xy" from the right hand side to the left side, but it seems to make the problem more confusing. Once I find the derivative (dy/dx) I know you just plug in the (x,y) value, so that's no problem.

If anyone can help, I would appreciate it greatly!

UPDATE!

I think I may have made the problem easier.. still going to try it out though. What if you got rid of the fraction on the left hand side from the beginning? Would that make it easier?
 
Last edited:
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  • #2
Why not plug in the values for x and y and then solve for dy/dx?
 
  • #3
Im not too sure what you mean? You mean plug in x= -1 and y=-1 into the equation? How would you take the derivative of that though? If you plug in numbers for the variables, you will come out with a number... a constant. The derivative of a constant is just zero... I am pretty sure (though I could be wrong) you plug in (-1,-1) at the end.

In case you missed my update, I think i know what i may have been doing wrong. I am going to try to get rid of the fraction from the beginning before i differntiate...may make it easier to differentiate.
 
  • #4
I meant that you plug them in AFTER you implicitly differentiate that expression. They you'll be left with a bunch of constants and the some dy/dx's.
 
  • #5
Oh I see, I am sorry for the misunderstanding. Yea, that's what I am trying to do right now. Essentially, I am trying to get it to be (dy/dx)= (variables here)

I think I may have figured it out. After going through, I get -3/3 when I plug in (-1,-1) so, in other words I get -1 as my answer. Anybody try it and get this?

thanks for the reply too!
 

1. What is implicit differentiation?

Implicit differentiation is a method used to find the derivative of an equation that is not in the form of y = f(x). It is used when the equation contains both x and y variables and cannot be easily solved for y. It involves differentiating both sides of the equation with respect to x and treating y as a function of x.

2. How is implicit differentiation different from explicit differentiation?

Explicit differentiation is used when the equation is already in the form of y = f(x), making it easier to solve for the derivative. Implicit differentiation is used when the equation is not in this form and requires the use of the chain rule to find the derivative of the y variable.

3. Why is implicit differentiation useful?

Implicit differentiation is useful for finding the derivative of equations that cannot be easily solved for y. It is often used in applications where the dependent variable cannot be explicitly solved for, such as in implicit functions or implicit curves.

4. Can implicit differentiation be used for higher order derivatives?

Yes, implicit differentiation can be applied for higher order derivatives by differentiating the equation with respect to x multiple times. However, it may become more complicated as the number of variables and functions increase.

5. What are some common mistakes when using implicit differentiation?

Some common mistakes when using implicit differentiation include forgetting to apply the chain rule, mistaking constants for variables, and incorrectly differentiating inverse trigonometric functions. It is important to carefully keep track of all variables and use the correct rules of differentiation.

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