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### Course: AP®︎/College Calculus AB>Unit 3

Lesson 3: Implicit differentiation

# Implicit differentiation review

Review your implicit differentiation skills and use them to solve problems.

## How do I perform implicit differentiation?

In implicit differentiation, we differentiate each side of an equation with two variables (usually $x$ and $y$) by treating one of the variables as a function of the other. This calls for using the chain rule.
Let's differentiate ${x}^{2}+{y}^{2}=1$ for example. Here, we treat $y$ as an implicit function of $x$.
$\begin{array}{rl}{x}^{2}+{y}^{2}& =1\\ \\ \frac{d}{dx}\left({x}^{2}+{y}^{2}\right)& =\frac{d}{dx}\left(1\right)\\ \\ \frac{d}{dx}\left({x}^{2}\right)+\frac{d}{dx}\left({y}^{2}\right)& =0\\ \\ 2x+2y\cdot \frac{dy}{dx}& =0\\ \\ 2y\cdot \frac{dy}{dx}& =-2x\\ \\ \frac{dy}{dx}& =-\frac{x}{y}\end{array}$
Notice that the derivative of ${y}^{2}$ is $2y\cdot \frac{dy}{dx}$ and not simply $2y$. This is because we treat $y$ as a function of $x$.
Want a deeper explanation of implicit differentiation? Check out this video.

Problem 1
${x}^{2}+xy+{y}^{3}=0$
$\frac{dy}{dx}=?$

Want to try more problems like this? Check out this exercise.

## Want to join the conversation?

• Hi !
i dont understand why in PROBLEM 1 :
2x + y + *bold*x dy....
dx
why x dy
dx

thank you!
• It's because of the product rule.

x*y differentiate into (1 (from differentiating the x))* (y) + (x) * (dy/dx (from differentiating the y)) = y + x*dy/dx
• Hi everyone, I have a quick question. We use the chain rule to differentiate "y^2" because we treat variable y as a function of x. However, when we have simple "y", we do not apply the chain rule and just express it as dy/dx. What is the difference between y^2 and y? Why to use chain rule in first case and not in the second one like 1(y(x))*dy/dx?
• We already know how to represent the derivative of y with respect to x: dy/dx, which is the thing we wish to find - in terms of x and y.

y² is a function of x AND of y.
Whenever we have a function of y we need to use the chain rule:
d/dx [ f(y) ] = d/dy [ f(y) ] · dy/dx

If it makes you feel easier we could say a 'simple *y"' is the identity function: f(y) = y.
Then d/dx [ f(y) ] = d/dy [ f(y) ] · dy/dx = dy/dy · dy/dx = 1 · dy/dx
• For problem 1, shouldn't it simplify to (-2x-y)/(x+3y^2)
• Yes it could, because (-2x-y) = -(2x+y). So you just distributed the minus one into the numerator, while it's left outside in the solution. It could also have been distributed into the denominator, since -1 = 1/-1.
• Why we can take derivative with respect to x or y both sides during implicit differentiation ?
• You learn in algebra that you can perform the same operation to both sides of an equation and the equation will still hold true. Taking a derivative just happens to be one such operation.
• Am I allowed to simplify an equation before doing implicit differentiation? Here is the question I was stuck on: y^2 = (x-1)/(x+1). When I attempt implicit differentiation the way it is and compare the answer to when I simplify the equation to (x+1)y^2= x-1, I got different answers.
• Excellent question!

Implicitly differentiating the original equation eventually yields dy/dx = 1/(y(x+1)^2).

Implicitly differentiating the simplified equation eventually yields dy/dx = (1-y^2)/(2y(x+1)).

So we compare 1/(y(x+1)^2) to (1-y^2)/(2y(x+1)), using y^2 = (x-1)/(x+1).

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

So the answers are really the same! The point is that the original equation or an equivalent form of this equation must be considered when comparing answers.

Have a blessed, wonderful day!
• Find the equation of the tangent line to the graph of the following equation at the point (-1,2) Implicit Differentiation

x^2 y - y^3 = 6x
• => y(2x) + (x^2)(dy/dx) - 3(y^2)(dy/dx) = 6
=> dy/dx = (6 - 2xy) / (x^2 - 3y^2)
• Hi everyone!

Do you happen to know any tricks and tips for solving derivatives and limits? Especially for implicit differentiation? I just don't like how long the process is taking me because I am a bit slow at writing and for our exams we have to write and it is time consuming.

Any help is much appreciated. Thank you!
• A short cut for implicit differentiation is using the partial derivative (∂/∂x). When you use the partial derivative, you treat all the variables, except the one you are differentiating with respect to, like a constant. For example ∂/∂x [2xy + y^2] = 2y. In this case, y is treated as a constant. Here is another example: ∂/∂y [2xy + y^2] = 2x + 2y. In this case, x is treated as the constant.

dy/dx = - [∂/∂x] / [∂/∂y] This is a shortcut to implicit differentiation.

Partial derivatives are formally covered in multivariable calculus.

Even though this is a multivariate topic, this method applies to single variable implicit differentiation because you are setting the output to be constant.

Hope this helps!
• I keep forgetting and recognizing that dy/dx(xy)=x'y+xy'. What lessons or video would be good for learning to recognize patterns in math like these, specifically ones having to do with Calculus?

Here's a summary:
   ------- (df * g)   |-----| |f  |     | | (f * dg)   |-----| |      g

d/dx(fg) = f' g + f g'
• Guys.. I have a question that's been bugging me:
Find the derivative function dy/dx of
x³y³-3x²y+3y²=5xy
Trying to confirm my amswer.