# How to differentiate 2x/(x+y) = y, implicitly?

• gillgill
In summary, implicit differentiation involves finding the derivative of an equation where the dependent variable is not explicitly written. The first method involves using the chain rule and the second method involves using the product rule. In both methods, the derivative of a constant is 0. The third question involves using the product rule with 3 terms, where one of the terms is a constant. The fourth question also involves using the product rule with 2 terms, but a sign error was made.
gillgill
differentiate
2x/(x+y)=y

Method1:
2x=xy+y^2
d/dx(2x)=d/dx(xy+y^2)
2=y+xy'+2y(y')
y'=(2-y)/(x+2y)

Method2:
2x/(x + y) = y
y' = [2(x + y) - (2x)(1 + y')]/(x + y)^2
[(x + y)^2 + 2x]y' = 2(x + y) - 2x
y' = 2y/[(x + y)^2 + 2x]

Which method is using "implicit" differentiation?

Both are, but the second is a lot sloppier.

another question:
x^3+y^3=6xy

3x^2+(3y^3)y'=...?
how do u differentiate 6xy?

product rule

gillgill said:
how do u differentiate 6xy?
Product rule!

i only learned how to differentiate using prodct rule with two terms...is it the same with 3 terms?

6 is a constant, don't even worry about it.

so...(6xy')+(6y)...?

6xy

u(x) = 6x, v(x) = y

(u(x)v(x))' = u'(x)v(x) + v'(x)u(x)

did i start this right?
x^2+y^2+2x-4y-20=0
2x+(2y)(y')+2+4y'=0

did i start this right?
x^2+y^2+2x-4y-20=0
2x+(2y)(y')+2+4y'=0

you seem to have made a sign error, but other than that it's okay.

o..i see it...thanks

## 1. How do you differentiate 2x/(x+y) = y implicitly?

To differentiate implicitly, you need to use the chain rule. First, you need to rewrite the equation as 2x(x+y)^-1 = y. Then, take the derivative of both sides with respect to x, treating y as a function of x. The derivative of y is dy/dx, and the derivative of x+y is 1, so the left side becomes 2x(-1)(x+y)^-2 (dx/dx) = dy/dx. Simplify the left side to get the final answer: dy/dx = -2x/(x+y)^2

## 2. What is the purpose of differentiating implicitly?

Implicit differentiation is used to find the derivative of a function that is not explicitly written in terms of x. This technique is useful when the dependent variable is not isolated on one side of the equation, making it difficult to take the derivative using traditional methods.

## 3. Is it possible to solve for y using implicit differentiation?

Yes, it is possible to solve for y using implicit differentiation. After taking the derivative, you can rearrange the equation to solve for dy/dx. Then, you can use algebraic manipulations to isolate y on one side of the equation. However, this may not always be possible, depending on the complexity of the equation.

## 4. Can implicit differentiation be used for any type of function?

Implicit differentiation can be used for any type of function, as long as it is possible to take the derivative of both sides of the equation. However, it may not always be the most efficient method for finding the derivative. In some cases, it may be easier to use other techniques such as the power rule or product rule.

## 5. What are some real-world applications of implicit differentiation?

Implicit differentiation is commonly used in physics and engineering to find rates of change and slopes of curves. It is also used in economics to analyze the relationship between variables. In addition, implicit differentiation is used in calculus to find the second derivative of a function.

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