# Implicit differentiation - Help

• pat666
In summary, the conversation is about finding y' given the equation x^3+y^2+x+y=2. The attempt at a solution involves taking the derivative and setting it equal to zero, but there is a mistake in setting up the equation. Eventually, the correct solution is found by factoring y' out of the equation and dividing both sides by -1-2y.
pat666

## Homework Statement

find y' given x3+y2+x+y=2

## The Attempt at a Solution

dy/dx=3x2+2y(y')+1+y'
I know that's wrong because my 89 gives it as y'=(-3x2+1)/(2y+1), I don't know how to get there though and what happened to the =2.
Thanks

## The Attempt at a Solution

dy/dx=3x2+2y(y')+1+y'
I know that's wrong because my 89 gives it as y'=(-3x2+1)/(2y+1), I don't know how to get there though and what happened to the =2.
Thanks

You turned the 2 in the original equation to dy/dx, instead of d(2)/dx, but the rest is correct. You should have 3x2+2y(y')+1+y'=0, then put the non-y terms on the right and factor out y' from 2y(y')+y' and you'll have the book's answer.

It also seems like the correct answer should have (-3x2-1) rather than (-3x2+1).

I've gotten closer to the solution but now I'm stuck at
3x^2+1=1-1-y'-2y(y')
3x^2+1=-y'-2y(y')?

Thanks

Now factor y' out of -y'-2y(y'). You'll have (-1-2y)y'. Then divide both sides by -1-2y.

Another thing..
in your last reply, you have =1-1-y'-2y(y'). Where did the 1-1 come from? If this is d(2)/dx, then it's sort of incorrect. The derivative of a constant is always 0; 1-1 is of course equal to zero, but writing 1-1 is sort of confusing for the person who's error checking.

yeah, I see what you mean - ill avoid doing that... Thanks for your help.

## What is implicit differentiation?

Implicit differentiation is a mathematical technique used to find the derivative of an implicit function, where the dependent and independent variables are not explicitly stated. It is most commonly used when the function is not easy to solve for one variable in terms of the other.

## What is the difference between implicit and explicit differentiation?

Explicit differentiation is used when the function is written in the form of y = f(x), where y is the dependent variable and x is the independent variable. Implicit differentiation is used when the function is not explicitly written in terms of y and x, making it more difficult to find the derivative.

## How do you perform implicit differentiation?

To perform implicit differentiation, you will need to use the chain rule and the product rule. First, take the derivative of both sides of the equation with respect to the independent variable. Then, use the chain rule to differentiate the dependent variable with respect to the independent variable. Finally, use the product rule to differentiate any terms with multiple variables.

## Why is implicit differentiation useful?

Implicit differentiation is useful because it allows us to find the derivative of a function that is not easily solved for one variable in terms of the other. It is often used in calculus to solve problems involving curves and surfaces, where the relationship between variables is not explicitly stated.

## What are some applications of implicit differentiation?

Some common applications of implicit differentiation include finding the slope of a tangent line to a curve, finding the rate of change in a system, and solving optimization problems. It is also used in physics, economics, and engineering to model and analyze real-world situations involving changing variables.

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