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Math 104: Calculus13 chapters | 104 lessons
Erin has taught math and science from grade school up to the post-graduate level. She holds a Ph.D. in Chemical Engineering.
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.
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:
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:
We end up with y` as some function of x and y.
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