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Implicit Differentiation: Examples & Formula

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  • 0:04 What Is Implicit…
  • 1:00 Examples
  • 3:27 Some Simple Guidelines
  • 4:23 Lesson Summary
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Lesson Transcript
Instructor
Jason Furney

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

Expert Contributor
Simona Hodis

Simona received her PhD in Applied Mathematics in 2010 and is a college professor teaching undergraduate mathematics courses.

This lesson takes you through the method of implicit differentiation. 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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ydy/dx


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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However, when we are finding dy/dx, we are taking the derivative with respect to x. (Every time you have taken the derivative the explicit way, you have been using this same method.)

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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Notice how every term that had a y now has a dy/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 factor.


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Additional Activities

Implicit Differentiation:

Review Key Terms

  • An implicit form of an equation of two variables is an equation where both variables are non-linear, so we cannot obtain an explicit form, where we write one variable in terms of the other.

For example, the equation of a circle in a rectangular equation is given as an equation where both variables appear squared, so it cannot be written explicitly as a single equation.

  • Implicit differentiation is the process of differentiation of an implicit form, where we make use of the Chain rule because the two variables are not linear.

The Chain rule applied to differentiate f(g(x)) is:

Implicit Differentiation - Related Rates

Implicit differentiation is commonly used in finding the slope of the tangent line to a curve given in rectangular form as an implicit form or in related rates problems.

In a related rate problem, we are given a rate of change of a variable and we need to find the rate of change of another variable.

For this, we need to do the following two steps:

  1. determine a relationship between the two variables, which usually is an implicit form
  2. use implicit differentiation to find a relationship between the two rates and solve for the unknown rate

Applications

Assuming you are flying a kite that maintains a constant height of 40 meters and the kite string is a straight line. If you are running to the right with 1 meter per second and the kite string is lengthening at 1 meter per second, to find the horizontal velocity of the kite at the instant when the kite is 30 meters behind you, do the following steps.

1) Denote the string displacement with x and the horizontal displacement of the kite with y. Obtain a relationship between x and y using the right triangle with the hypotenuse, 50 + x, and the sides with a vertical length of 40 and the horizontal length y + 30 + x.

2) Use implicit differentiation to differentiate the relationship in 1) and find y'.

Solutions

1)

2) Using the instant x = 0, y = 0, x' = 1, y' = 1, we obtain:

So the kite moves horizontally to the left at a rate of 0.67 meters per second.

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