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Implicit Differentiation: Examples & Formula

Instructor: Jason Furney

Jason has taught both College and High School Mathematics and holds a Master's Degree in Math Education.

This lesson will take you through the method of implicit differentiation with multiple examples and a quiz at the end to test your knowledge. Implicit differentiation utilizes all of your basic derivative rules to find the derivative of equations that are not in standard form.

What Is Implicit Differentiation?

Up to this point in calculus, most functions that have been derived were in explicit form.

Explicit form is the standard y = 2x + 5 or any other function where y is on one side of the equal sign and x is on the other!

Implicit Functions are different in that x and y can be on the same side.

A simple example is: xy = 1. It is here that implicit differentiation is used.

Remember! You HAVE used all of these derivative rules before! y has just been isolated for you.

What happens when you take the derivative of y = 2x? Let's take a look:

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y

Every time, we take the derivative of a y variable, we will write either y' or dy/dx.

REMEMBER: All other derivative rules still apply.

Examples

Ex 1: Find dy/dx given that y^3 + 2y^2 - 3y +x^2 = -2 (Written out in the image below)

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To begin, let's take the derivative of y^3. The 3 moves down in front of the y and the exponent decreases by 1... just like our standard derivative!

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dy/dxx

In this case, our variable is a y so we have to go one step further. That step is the chain rule where f(x) = x^3 and g(x) = y.

Since we have already taken the derivative of f(x)... we need to find the derivative of g(x) or just y. The derivative of y is dy/dx.

The chain rule tells us the derivative is f '(g(x)) * g'(x) or:

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That's great! Now we can use the same method to derive the rest of the y variables.

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ydy/dx

To finish taking the derivative, we have to find dy/dx of x^2 and of -2. Here, everything is normal.

dy/dx of x^2 = 2x and dy/dx of -2 = 0. Putting all of our individual derivatives together, we get:

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But are we done? The answer to that is NO. To finish properly taking a derivative implicitly, we need to solve for dy/dx. To do this, treat dy/dx like it is just another variable and use your algebra skills to isolate it and solve for it.

First: Subtract anything that does not have dy/dx to the other side of the equal sign.

Second: Factor out dy/dx.

Third: Divide both sides by what remains after you factored.

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Derive -> Isolate -> Solve

Ex: 2 Find dy/dx given xy - y = 3

Any time you have two variables touching, you need to use the Product Rule

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