How to Find Derivatives of Implicit Functions

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  • 0:06 Implicit Function
  • 1:53 Derivatives of an…
  • 4:54 Implicit Differentiation
  • 8:34 Lesson Summary
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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)!

Implicit Function

Diagram of the land owned by Uncle Joe
Uncle Joes Land

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 2x + 2y; 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(2x + 2y) = 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.

Derivatives of an Implicit Function

Finding y prime in the Uncle Joe example
Implicit Function Derivatives Example 1

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!

Steps for Implicit Differentiation

In the final example, use the product rule on the first term, ye^x
Implicit Function Derivatives Example 2

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

  1. Differentiate both sides.
  2. Collect y` terms to one side of the equation.
  3. Factor y` out of the terms.
  4. 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.

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