Implicit differentiation (advanced example)

Let's say we have the relationship y is equal to cosine of 5x minus 3y. And what I want to find is the rate at which y is changing with respect to x. And we'll assume that y is a function of x. So let's do what we've always been doing. Let's apply the derivative operator to both sides of this equation. On the left-hand side, right over here, we get dy/dx is equal to-- now here on the right hand side, we're going to apply the chain rule. The derivative of the cosine of something, with respect to that something, is going to be equal to negative sine of that something. So negative sine of 5x minus 3y. And then we have to multiply that by the derivative of that something with respect to x. So what's the derivative of the something with respect to x? Well the derivative of 5x with respect to x is just equal to 5. And the derivative of negative 3y with respect to x is just negative 3 times dy/dx. Negative 3 times the derivative of y with respect to x. And now we just need to solve for dy/dx. And as you can see, with some of these implicit differentiation problems, this is the hard part. And actually, let me make that dy/dx the same color. So that we can keep track of it easier. So this is going to be dy/dx. And then I can close the parentheses. So how can we do it? It's just going to be a little bit of algebra to work through. Well, we can distribute the sine of 5x minus 3y. So let me rewrite everything. We get dy-- whoops, I'm going to do that in the yellow color-- we get dy/dx is equal to-- you distribute the negative sine of 5x minus 3y. You get-- so let me make sure we know what we're doing. It's going to be, we're going to distribute that, and we're going to distribute that. So you're going to have 5 times all of this. So you're going to have negative this 5 times the sine of 5x minus 3y. And then you're going to have the negative times a negative, those are going to, you're going to end up with a positive. And so you're going to end up with plus 3 times the sine of 5x minus 3y dy/dx. Now what we can do is subtract 3 sine of 5x minus 3y from both sides. So just to be clear, this is essentially a 1 dy/dx. So if we subtract this from both sides, we are left with-- So on the left-hand side, we're going to have a 1 dy/dx, and we're going to subtract from that 3 sine of 5x minus 3y dy/dx's. So you're going to have 1 minus 3-- I'll keep the color for the 3 for fun-- 3 sine of 5x minus 3y dy/dx's on the left-hand side, is going to be equal to, well, we subtracted this from both sides. So on the right-hand side, this is going to go away. So we're just going to be left with a negative 5 sine of 5x minus 3y. And we're in the home stretch now. To solve for dy/dx, we just have to divide both sides of the equation by this. And we are left with dy/dx is equal to this thing, negative 5 times the sine of 5x minus 3y. All of that over 1 minus 3 sine of 5x minus 3y. And we are done.