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## Calculus 1

### Course: Calculus 1>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, squared, plus, y, squared, equals, 1 for example. Here, we treat y as an implicit function of x.
\begin{aligned} x^2+y^2&=1 \\\\ \dfrac{d}{dx}(x^2+y^2)&=\dfrac{d}{dx}(1) \\\\ \dfrac{d}{dx}(x^2)+\dfrac{d}{dx}(y^2)&=0 \\\\ 2x+2y\cdot\dfrac{dy}{dx}&=0 \\\\ 2y\cdot\dfrac{dy}{dx}&=-2x \\\\ \dfrac{dy}{dx}&=-\dfrac{x}{y} \end{aligned}
Notice that the derivative of y, squared is 2, y, dot, start fraction, d, y, divided by, d, x, end fraction and not simply 2, y. This is because we treat y as a function of x.
Want a deeper explanation of implicit differentiation? Check out this video.

Problem 1
• Current
x, squared, plus, x, y, plus, y, cubed, equals, 0
start fraction, d, y, divided by, d, x, end fraction, equals, question mark

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 ?
(1 vote)
• 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.
• 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)
(1 vote)
• 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!
• 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!
• does someone know the derivative of cosx=2y-xy?
• differentiating both sides with respect to x:
-sinx=2(dy/dx)-(y +x(dy/dx))
• 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'
• do you treat dy/dx like y? For example, if y is cubed and muliplied by x, is dy/dx?