Learn about More implicit differentiation

Khan Academy Presents: 2 more implicit differentiation examples

Learn about More implicit differentiation I’ve been ask to do a few more implicit differentiation problems and this is something that I think confuses a lot of people. So let’s do a couple more of problems and this is—I think there’s a type of—I forgot the user that actually sent this but I’m guessing this is what he intended and if this is what he intended this is something close so it should give the idea. - So the implicit function they gave was x² - 24xy + 16y²-400x-300y = 0. So how do we do this? Well we take the derivative of both sides with respect to x. I could write the derivative operator on both sides but I think you know that at this point so that’s for next step. So let’s do that and I’ll do it in blue. Okay derivative of x² with respect to x is 2x and now let’s see its going to be -24. I’m going to take the constant term out. So 2x – 24(1 x y + x and this is sometimes confuses people because it’s so easy. Well just dy dx and instead of writing dy dx I’m just going to write y prime here just to get you use to a different type of notation although I preferred dy dx because you realize it you’re using the chain rule but anyway 2x-24(1 y +xy’) + 32y but we’re taking the derivative with respect to x so we have to take the derivative of this with respect to y which we did and then multiply that times the derivative of y with respect to x which is dy dx or y prime and this with respect to x well that’s just minus 400 and then what’s the derivative of this with respect to x let’s minus 300 and then times the derivative of y with respect x because were doing the chain rule right. Minus 300 is just the derivative of this with respect to y and then I want to hit this point. Well you know—some people and they learn just say oh whenever I take a derivative of y just take the derivative and I’ multiply by y prime and that’s right, that’s essential what you do but I want you to understand that you’re doing the chain rule when you do that. So you take the derivative with respect to y and then multiplying that times the derivative with respect to x which will right is y prime now and all of that is equal to zero. Now let’s simplify this. So let’s see I get 2x -24y – 24xy’ + 32yy’ – 400 – 300y’ = 0 and let’s switch colors. So let’s separate all the y prime terms out. So let’s see this is one of the prime terms. We have this term as the y’ in it. We have this term has a y prime in it and we have this term as the y prime in it. So we can write this, let’s do the positive one first. So 32y, so this is the same thing as y’(32y – 24x – 300) + 2x – 24y – 400 = 0. I just rearrange and took out the y prime out of this, this and this term. let’s take all of these to the other side equation. Now essentially what I want to do is get all the terms that don’t have the y prime unto the right hand side and then divided by this and then we will solve the equation and let see. So we get y’ times and this is you can also write you know the derivative with respect to x or dy dx. In general even though it takes more time I like to normally write dy dx because you remember I’m taking the derivative with respect to x. Y prime I don’t know. You might be thinking the derivative with respect to who knows what. So let’s see y’(32y-24x-300) is equal to put all of this on the right hand side so you get I’ll switch the one like put the positive numbers first. 24y – 2x + 400 and then divide both sides by this and I’ll do it in different color because we’re almost at the answer and we get y’ the derivative with respect to x is equal to this thing y’ = 24y – 2x + 400 all of that over this thing 32y - 24x – 300 and let’s see we could at least it looks like well get everything this little by 2 so we could divide the numerator and the denominator by two and you get 12y – x + 200 over 16y – 12x – 150 and if I didn’t make a careless mistake that’s the answer. The derivative of y with respect to x anytime that this expression that has y and x in it and that’s cool because this is a fairly convoluted implicit function but we can figure out the slope at any point because we know the derivative. Let’s do the next one. This one looks simpler (x – y)² = 8(y – 6). All right so let’s just take the derivative with respect to x on both sides and I’ll use the dy dx notation this time. So let’s first use the chain rule. Let’s take the derivative of the inside. So what’s the derivative of the inside function? It is—well we could do the either way. Let’s do the other one. Let’s do inside and out but let’s take the derivative of this entire expression with respect to x – y, well that is 2(x – y) and then its going to be that times the derivative of x – y with respect to x. So what’s the derivative of x – y with respect to x. Well the derivative of x with respect to x is just one and what’s the derivative of well we can say – y with respect to x. let’s just minus dy dx. You could say the derivative of y with respect to y is the derivative of minus y with respect to y is minus one and then by the chain rule multiply that times derivative of y with respect to x either way until you get the point hopefully. So that’s the derivative of the left hand side and what is the derivative of the right hand side? Well we could do the chain rule or whatever but you know this is the same thing as 8y – 48 right. So we could rewrite this as 8y – 48 and so if you take the derivative of this with respect to x this is the constant doesn’t contribute so what’s the derivative of 8y with respect to x? Once the derivative of 8y with respect to y which is just 8 and by the chain rule times the derivative of 8y with respect to y times the derivative of y with respect to x. and now we just solve for the dy dx. So let’s see you get and I will do a different color now. (2x - 2y)(1-dy/dx) = 8 dy/dx and then what can we do? We can distribute this out so 2x times this whole thing so 2x - 2x dy/dx right that times that and then that times that and then this times this so minus 2y and then minus minus so plus 2y dy/dx and then let’s take this because I want to put all the dy dx terms on the side. So let’s subtract this from both side of this equation so then we get -8 dy/dx = 0 and then the next step, well let just separate on all the dy dx terms so that’s the dy dx term, that’s one, that’s one there you can see the spot and when do you write y’ So let’s see so if we take those together and take out the dy dx we get dy the derivative of y with respect to x times then you see any positive you’ll see that one first. 2y – 2x right that’s this one minus 8 and then plus 2x – 2y = 0. Let’s put these onto the right hand side of the equation. Were almost there dy/dx (2y-2x-8) = 2y – 2x let’s just put on the right hand side and now divide both sides by this and we are done. Dy/dx = 2y – 2x over 2y – 2x – 8 and once again we can divide the numerator and the denominator by 2. So were left with y – x over y - x – 8 and were done. Hopefully you found that helpful.