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Math 104: Calculus14 chapters | 116 lessons | 11 flashcard sets

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Lesson Transcript

Instructor:
*Robert Egan*

How do you define the rate of change when your function has variables that cannot be separated? Learn how implicit differentiation can be used to find dy/dx even when you don't have y=f(x)!

I have an Uncle Joe who's a farmer. He really likes math, so he told me about his plot of land. He said his land extends *x* meters to the east and *y* meters to the north. He said, 'You know, the area of that land I have is (*x*)(*y*).' That's because his land is just a rectangle, and the border, or perimeter, around his land is just 2*x* + 2*y*; we've got *x* + *y* + *x* + *y*. Now Uncle Joe told me that his land always satisfies one condition. That is that the area of his land is always equal to half of the perimeter. In other words, (*x*)(*y*) = 1/2(2*x* + 2*y*) = *x* + *y*. That's great, Uncle Joe! You love math - what do you need me for?

Well, Uncle Joe always wants this equation, (*x*)(*y*) = *x* + *y*, to be true. He's in the land business. He wants to know if he buys more land to the east - so if he changes *x* - how much does he have to change *y* to keep this equation true? He wants to know *dy/dx* - how much *y* should change while *x* is changing. Oh, so Uncle Joe wants me to calculate a derivative. I can do this.

Okay, find *dy/dx* for (*x*)(*y*) = *x* + *y*. The first thing I want to do is set up *y*=*f(x)* ... uh-oh, I can't do that; I can't separate *x* and *y* to different sides of this particular equation. This is going to make Farmer Joe really unhappy. Well, maybe there's another way to find *dy/dx*. I remember that to find *dy/dx* of *f(x)*, I wrote *y*=*f(x)* and differentiated both sides. I got *dy/dx* = *d/dx f(x)*. Why don't I try that here?

*d/dx*(*xy*) = *d/dx*(*x* + *y*). For *d/dx*(*xy*), that looks like something I need to use the product rule on: *d/dx*(*x*)*y* + *d/dx*(*y*)*x*. The first term is 1 * *y*, or *y*, plus *dy/dx*(*x*). I can write *dy/dx* as *y`*, then the whole left-hand side of my equation becomes *y* + *xy`*. The right-hand side of my equation is *d/dx*(*x* + *y*). I can split that up and write *d/dx*(*x*) + *d/dx*(*y*). That's just 1 + *dy/dx* = 1 + *y`*. My entire equation - when I differentiate (*x*)(*y*) = *x* + *y* - is *y* + *xy`* = 1 + *y`*. Alright, but I'm trying to find *dy/dx*, or *y`*.

So let's solve this for *y`*. First, let's collect all the terms, move all the *y`* terms to the left-hand side, *xy`* - *y`* = 1 - *y*. Let's factor out *y`*, so I have *y`*(*x* - 1) = 1 - *y*. Then let's divide the left- and right-hand sides by (*x* - 1), and we end up with *y`* = (1 - *y*) / (*x* - 1). Whew, that was a lot of work for Uncle Joe!

What we just did is an example of **implicit differentiation**. For implicit differentiation, we can follow these steps:

- Differentiate both sides.
- Collect
*y`*terms to one side of the equation. - Factor
*y`*out of the terms. - Solve for
*y`*.

Let's do an example. Let's say we have *y* = *ye*^*x* + *x*, and let's say we're trying to find *y`*, or *dy/dx*.

Our first step is to differentiate both sides. *d/dx*(*y*) = *d/dx*(*ye*^*x* + *x*). So *d/dx*(*y*) is just *dy/dx*. *d/dx* of the right-hand side is a little bit more complicated. Let's look at this first term, *ye*^*x*. The derivative of *ye*^*x* is *y*(*d/dx*(*e*^*x*)) + *e*^*x*(*d/dx*(*y*)). That's just using the product rule. This first term becomes *ye*^*x* + *e*^*x* *dy/dx*. The second term is just *x*, and if we take the derivative of *x* with respect to *x*, we get 1. Alright, so let's write *dy/dx* as *y`*: *y`* = *ye*^*x* + *e*^(*x*)*y`* + 1. Great, we've differentiated both sides.

Now, let's collect all of the *y`* terms on one side of the equation: *y`* - *e*^(*x*)*y`* = *ye*^*x* + 1. So all I've done is move *e*^(*x*)*y`* to the left-hand side. Fantastic, halfway there. Now I'm going to factor out *y`*: *y`*(1 - *e*^*x*) = *ye*^*x* + 1. Finally, I'm going to solve for *y`* by dividing everything by (1 - *e*^*x*), so *y`* = *ye*^*x* + 1 / (1 - *e*^*x*). This is great, as long as *x* doesn't equal zero. If *x*=0, we're trying to divide by zero, and we can't do that.

This is how we find *y`*, or *dy/dx*, for another case where we have an implicit equation, *y* = *ye*^*x* + *x*.

**Implicit differentiation** is what you use when you have *x* and *y* on both sides of an equation and you're looking for *dy/dx*. We did this in the case of Farmer Joe's land when he gave us the equation (*x*)(*y*) = *x* + *y*.

To do implicit differentiation, we:

- Differentiate both sides.
- Collect
*y`*terms to one side of the equation. - Factor
*y`*out of those terms. - Solve for
*y`*.

We end up with *y`* as some function of *x* and *y*.

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Math 104: Calculus14 chapters | 116 lessons | 11 flashcard sets

- Go to Continuity

- Go to Limits

- Using Limits to Calculate the Derivative 8:11
- The Linear Properties of a Derivative 8:31
- Calculating Derivatives of Trigonometric Functions 7:20
- Calculating Derivatives of Polynomial Equations 10:25
- Calculating Derivatives of Exponential Equations 8:56
- Using the Chain Rule to Differentiate Complex Functions 9:40
- Differentiating Factored Polynomials: Product Rule and Expansion 6:44
- When to Use the Quotient Rule for Differentiation 7:54
- Understanding Higher Order Derivatives Using Graphs 7:25
- Calculating Higher Order Derivatives 9:24
- How to Find Derivatives of Implicit Functions 9:23
- Applying the Rules of Differentiation to Calculate Derivatives 11:09
- Optimization Problems in Calculus: Examples & Explanation 10:45
- Go to Calculating Derivatives and Derivative Rules

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