Okay in this example, we're gonna talk about implicit differentiation. And implicit differentiation is really nothing much different than regular differentiation. First of an explicit equation is ones that, if I can spell correctly, is one where you normally have y by itself and x is on the other side.
So these are 2 examples of explicit equations. An implicit equation is one where the x's and y's are all mixed up on the same side.
So you can see we have an x and a y on the same side that's considered an implicit equation.
So that is all the terminology means right there. The main idea to remember is normally we take the derivative with respect to x. For example with x squared and for that we simply get 2x.
Notice that these symbols match up. Well if I take the derivative with respect to x, say of y squared the symbols don't match up.
And to take the derivative we would like them to. So the idea is we'll simply rewrite the d/dx. Imagine as it being a fraction as d/dy.
dy/dx and we still have our y squared hanging out there. So right imagine the dy's canceling out and I would have d/dx again. Well I can put the y square inside of the 2.
And now the symbols do match up. So if I take the derivative of this part I'm gonna get 2y. But notice there's a dy/dx left hanging out.
And this is gonna be the moral of the story when doing implicit differentiation. Basically take the derivative like normal, and if you're taking the derivative of something say involving a y or a z.
As long are you're taking the derivative with respect to x you'll tack on a dy/dx or a dz/dx. Okay so in general the procedure for a doing one of these implicit differentiation examples bsically just take the derivative, like normal.
Again, adding dy/dx' s where needed. And you only need those when you're taking the derivative of something not involving x. The next thing just get rid of any parentheses that may be floating around. And lastly just solve that equation for dy/dx.
So let's do a basic, a basic example here. Got one made out. So we want to find dy/dx and notice I do have an implicit equation. I've got some x's and y's jumbled up here.
So if I take the derivative of this thing, again the derivative of x squared, well that's just 2x. And I'm gonna have to use the product rule on this next part because I have a product x times y.
So if I take the derivative of x, I just get one. I'll leave the y alone. I don't need a dy/dx because I didn't take the derivative of the y piece yet.
Plus now I'll leave the x alone. And the derivative of y is one dy/dx. Next we take the derivative of cosine of y. That'll give us negative siny.
And again I have to tack on my dy/dx. And on the right hand side the derivative of 8y will be 8 dy/dx.
Okay so now I've done the derivative part. The only thing really left that I have to do in this case is to get the dy/dx' s on one side and everything else to the other.
So, I'm just gonna remove my parentheses here. Normally you would distribute if there was anything to distribute but in this case there's reall not. So get rid of your parentheses.
dy/dx and that equals 8 dy/dx. Okay so notice I have 3 terms that have dy/dx 's and Ihave 2 terms that have no dy/dx 's.
So I'm gonna keep the terms that have a dy/dx on the left side. So I'll have x dy/dx minus siny dy/dx. I'm gonna move the 8 dy/dx over by subtracting it.
So minus 8 dy/dx. And now the 2 terms on the left, the 2x and the 2y that didn't have a dy/dx, I'm gonna subtract those over.
So minus 2x will bring the 2x over, and minus y will bring the y over. Now the point of doing this is everything on the left has a dy/dx in it so I can factor that out.
So I'll pull out the dy/dx and I'm left with x minus siny minus 8. So I just pull out the dy/dx, the dy/dx, the dy/dx.
There's nothing left to do on the right hand side. And to get dy/dx by itself we simply divide by all of the stuff in the parentheses.
So I'll be left with dy/dx on the left side and on the right side I'll be left with negative 2x minus y. And after dividing I'll have x minus siny minus 8.
And that will be our solution to this problem. So again the basic idea all you really need to remember when you're doing one of these is when you take the derivative
involving a y term do it like normal but simply tack on your dy/dx and it's certainly something easy to kind of leave out and forget, especially
if you're doing these for the first time. But other than that that's the only real trick
The algebra can get a little tedious in some of them but like I said just move your dy/dx 's to one side, everything else to the other
factor out the dy/dx and divide by what's left over and that'll be your answer.