Let’s get some more practice

doing implicit differentiation. So let’s find the derivative

of y with respect to x. We’re going to assume

that y is a function of x. So let’s apply our

derivative operator to both sides of this equation. So let’s apply our

derivative operator. And so first, on

the left hand side, we essentially are just going

to apply the chain rule. First we have the

derivative with respect to x of x minus y squared. So the chain rule

tells us this is going to be the derivative

of the something squared with respect to the

something, which is just going to be 2 times x

minus y to the first power. I won’t write the

1 right over there. Times the derivative of the

something with respect to x. Well, the derivative of x

with respect to x is just 1, and the derivative

of y with respect to x, that’s what

we’re trying to solve. So it’s going to

be 1 minus dy dx. Let me make it a

little bit clearer what I just did right over here. This right over here

is the derivative of x minus y squared with

respect to x minus y. And then this right over

here is the derivative of x minus y with respect to x. Just the chain rule. Now let’s go to the right

hand side of this equation. This is going to be equal to

the derivative of x with respect to x is 1. The derivative of y

with respect to x. We’re just going to write

that as the derivative of y with respect to x. And then finally, the

derivative with respect to x of a constant, that’s

just going to be equal to 0. Now let’s see if we can

solve for the derivative of y with respect to x. So the most obvious thing to do. Let’s make it clear. This right over here, I

can rewrite as 2x minus 2y. So let me do that so

I can save some space. This is 2x minus 2y If

I just distribute the 2. And now I can distribute

the 2x minus 2y onto each of these terms. So 2x minus 2y times 1 is

just going to be 2x minus 2y. And then 2x minus 2y

times negative dy dx, that’s just going to be

negative 2x minus 2y. Or we could write that as

2y minus 2x times dy dx. Is equal to 1 plus dy dx. I’ll do all my dy

dx’s in orange now. 1 plus dy dx. So now there’s a

couple of things that we could attempt to do. We could subtract 2x

minus 2y from both sides. So let’s do that. So let’s subtract 2x

minus 2y from both sides. So over here, we’re going

to subtract 2x minus 2y from that side. And then we could also subtract

a dy dx from both sides, so that all of our dy dx’s

are on the left hand side, and all of our non dy dx’s

are on the right hand side. So let’s do that. So we’re going to subtract a

dy dx on the right and a dy dx here on the left. And so what are we left with? Well, on the left hand

side, these cancel out. And we’re left with 2y minus

2x dy dx minus 1 dy dx, or just minus a dy dx. Let me make it clear. We could write this

as a minus 1 dy dx. So this is we can

essentially just add these two coefficients. So this simplifies to 2y minus

2x minus 1 times the derivative of y with respect

to x, which is going to be equal to– on this

side, this cancels out. We are left with 1

minus 2x plus 2y. So let me write it that way. Or we could write

this as– so negative, negative 2y is

just a positive 2y. And then we have minus 2x. And then we add that 1, plus 1. And now to solve

for dy dx, we just have to divide both sides

by 2y minus 2x minus 1. And we are left

with– we deserve a little bit of a drum

roll at this point. As you can see, the

hardest part was really the algebra to solve for dy dx. We get the derivative

of y with respect to x is equal to

2y minus 2x plus 1 over 2y minus 2x minus 1.

Could you have solved for y before differentiating?

Please, explain – Is Sal redoing all his Calculus Videos or what?? What's the difference between these and previous old Calculus videos??

One of the best channels on youtube. Thank you is not enough. <3

HD quality 🙂

A bat and a ball cost $1.10 in total. The bat costs $1 more than the ball. How much does the ball cost?

Mistake at 2:17: It should be (2x – 2y) -(2y-2x)dx/dy