# Superior implicit differentiation, prove answer.

• Sakha
In summary, the given equation 2XY = Y^2 can be simplified to y''2 = (y^2-2xy)/(y-x)^3 by substituting the values of dy/dx and d^2y/dx^2. However, there may be an error in the given solution as the final expression does not match with the simplified equation.
Sakha

## Homework Statement

$$2XY = Y^2 prove that y''2 = \frac{y^{2}-2xy}{(y-x)^3}$$
EDIT: Sorry, don't know how to insert a space in Latex.

## The Attempt at a Solution

$$2y+2x \frac{dy}{dx} = 2y \frac{dy}{dx}$$
$$\frac{dy}{dx}(2y-2x) = 2y$$
$$\frac{dy}{dx}= \frac{y}{y-x}$$
$$\frac{d^2y}{dx^2}=\frac{\frac{dy}{dx}(y-x)-(\frac{dy}{dx}-1)y}{(y-x)^2}$$
$$\frac{d^2y}{dx^2}= \frac{y^2}{(y-x)^3}$$Is my work correct? If yes, then the question itself is wrong.
In the original equation (2XY = Y2) does it matters that the letters are caps?

Your work is correct up to the last line. But the given solution is the correct one. Looks like something went wrong when you substituted your expression for dy/dx in and simplified.

Plugging in gives me:
((y/y-x)(y-x)-((y/y-x)-1)y)/((y-x)2)
Simplifiying:
(y-(y2/(y-x))-y))/((y-x)2)
(-y2/(y-x))/((y-x)2)

Different than my first try (now it's negative), but still I don't know how to the the -2xy in there.

Sakha said:
Plugging in gives me:
((y/y-x)(y-x)-((y/y-x)-1)y)/((y-x)2)
Simplifiying:
(y-(y2/(y-x))-y))/((y-x)2)
(-y2/(y-x))/((y-x)2)

Different than my first try (now it's negative), but still I don't know how to the the -2xy in there.

In y-(y^2/(y-x)-y) the y's don't cancel. It's minus of a minus.

Got it, my mistake was that I moved the (y-x) that was in the numerator as a denumerator before adding fractions. And the wrong minus of a minus that you pointed.
Thanks.

## 1. What is implicit differentiation?

Implicit differentiation is a mathematical technique used to find the derivative of a function that is not expressed in the form of y = f(x). Instead, the function is expressed as an equation that relates x and y, and the derivative is found with respect to x.

## 2. How is implicit differentiation different from explicit differentiation?

Explicit differentiation is used to find the derivative of a function that is expressed in the form of y = f(x), where y is isolated on one side of the equation. Implicit differentiation is used for functions that cannot be easily solved for y and require the use of the chain rule.

## 3. What are the steps for performing implicit differentiation?

The steps for implicit differentiation are as follows:
1. Differentiate both sides of the equation with respect to x.
2. Use the chain rule to differentiate any terms with y in them.
3. Collect all terms with dy/dx on one side of the equation.
4. Solve for dy/dx.

## 4. Why is implicit differentiation useful?

Implicit differentiation is useful because it allows us to find the derivative of functions that cannot be solved for y in terms of x. This is important in many areas of mathematics and science, such as physics, engineering, and economics.

## 5. Can you provide an example of a problem where implicit differentiation is used?

One example of a problem where implicit differentiation is used is finding the slope of a tangent line to a curve that is not expressed in the form of y = f(x). This is often seen in physics problems involving motion, where the position of an object is given as a function of time, and the velocity (or slope of the tangent line) needs to be determined using implicit differentiation.

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