What Is the Correct Approach to Solve These Implicit Differentiation Problems?

In summary, the conversation is about differentiating the equation 8x^2-10xy+3y^2=26 and finding the correct answer, as well as a question about multiplying both sides of the equation by a common denominator. The correct answer is (5y-8x)/(3y-5x) and the reason for multiplying by (2x^1/2 * y^1/2) is because it is the common denominator for the terms in the equation.
  • #1
pr0blumz
66
0

Homework Statement


8x^2-10xy+3y^2=26


2. The attempt at a solution

(8)(2x)-(-10x)y'+(y)(-10)+(3)(2y)y'=0

16x+10x(y')-10y+6y(y')=0

y'(10x+6y)+16x-10y=0

y'(10x+6y)=10y-16x

y'=(10y-16x)/(10x+6y)

y'=(5y-8x)/(5x+3y)

I know I'm doing something wrong but I can't see it for myself. Can someome point me into the right direction?
 
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  • #2
pr0blumz said:
(8)(2x)-(-10x)y'+(y)(-10)+(3)(2y)y'=0

Where'd you get that extra minus sign from when differentiating -10xy?
 
  • #3
I differentiated (-10xy)
 
  • #4
pr0blumz said:
(8)(2x)-(-10x)y'+(y)(-10)+(3)(2y)y'=0

That -ve sign should be a + sign.
 
  • #5
I'm still getting (5y-8x)/(3y-5x)
 
  • #6
pr0blumz said:
I'm still getting (5y-8x)/(3y-5x)

What answer are you supposed to get?
 
  • #7
(8x-5y)/(5x-3y)
 
  • #8
pr0blumz said:
(8x-5y)/(5x-3y)

Well multiply both the numerator and denominator by -1.
 
  • #9
Only thing that I see happened was that y' was moved to the right side of the equation therefore changing the signs. After I done it that way, I got the correct answer. Go figure!
 
  • #10
As rockfreak667 said before, you already had the correct answer:
(5y-8x)/(3y-5x)= (8x-5y)/(5x-3y)
 
  • #11
HallsofIvy said:
As rockfreak667 said before, you already had the correct answer:
(5y-8x)/(3y-5x)= (8x-5y)/(5x-3y)

I have another problem as well.

x + (sqrtx)(sqrty) = 2y

1 + (x^1/2)/(2y^1/2)y' + (y^1/2)/(2x^1/2) = 2y'

My question is why I suppose to multiply both sides by (2x^1/2 * y^1/2) and not both of the denominators?
 
  • #12
pr0blumz said:
I have another problem as well.

x + (sqrtx)(sqrty) = 2y

1 + (x^1/2)/(2y^1/2)y' + (y^1/2)/(2x^1/2) = 2y'

My question is why I suppose to multiply both sides by (2x^1/2 * y^1/2) and not both of the denominators?

the common denominator of those the terms (not the 1 and not the 2y') is 2x1/2y1/2
 

FAQ: What Is the Correct Approach to Solve These Implicit Differentiation Problems?

What is implicit differentiation?

Implicit differentiation is a mathematical method used to find the derivative of a function that is not explicitly defined in terms of one variable. It is often used when the function is difficult or impossible to solve for one variable alone.

Why is implicit differentiation useful?

Implicit differentiation allows us to find the derivative of a function without having to solve for one variable. This can be helpful when dealing with complex functions or when the function is not defined explicitly in terms of one variable.

How is implicit differentiation different from explicit differentiation?

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

What are the steps for implicit differentiation?

The steps for implicit differentiation are as follows: 1) Differentiate both sides of the equation with respect to the variable of interest, treating the other variables as constants. 2) Simplify the resulting equation by collecting like terms. 3) Solve for the derivative by isolating the variable of interest on one side of the equation.

What are some real-world applications of implicit differentiation?

Implicit differentiation is used in many fields of science and engineering, such as physics, economics, and biology. It can be used to find the rate of change in a system or to optimize a process. For example, it can be used to find the maximum profit for a company or to determine the velocity of an object in motion.

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