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### Course: AP®︎/College Calculus AB > Unit 3

Lesson 3: Implicit differentiation- Implicit differentiation
- Worked example: Implicit differentiation
- Worked example: Evaluating derivative with implicit differentiation
- Implicit differentiation
- Showing explicit and implicit differentiation give same result
- Implicit differentiation review

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# Worked example: Implicit differentiation

Using implicit differentiation, let's take on the challenge of the equation (x-y)² = x + y - 1 in this worked example. We utilize the chain rule and algebraic techniques to find the derivative of y with respect to x. This adventure deepens our grasp of how variables interact within intricate equations. Created by Sal Khan.

## Want to join the conversation?

- This whole "with respect to y or respect to x" thing has me very confused. And not just bc of this video, but bc my really really expensive text book doesn't really talk much about this concept. I don't really understand the dy/dx notation. I always understood it to mean simply "derivative." But now it's a variable... What does "with respect to" mean? I know what an equation in explicit and implicit form means if that helps anyone explain my problem.(98 votes)
- I know it's a late reply but maybe people are still wondering. Let's say we have a function y=x^2. Derivative of y with respect to x simply means the rate of change in y for a very small change in x. So, the slope for a given x. If I have something like 'derivative of y with respect to x^2 then it means the rate of change in y for a very small change in x^2. So, the slope for a given value of x^2 (you plot x^2 on the x-axis in this case). We would write this second derivative as dy/d(x^2).(57 votes)

- Did Sal made a mistake? Should the answer should be the following?:

dy/dx = (2x - 2y - 1) / (2x - 2y + 1)

I've tried it twice and got the the same answer. Wolfram Alpha seems to confirm: http://www.wolframalpha.com/input/?i=derivative+with+respect+to+x+%28x-y%29%5E2+%3D+x+%2B+y+-+1.

Workings:

2(x - y)(1 - dy/dx) = 1 + dy/dx

=> (2x - 2y)(1 - dy/dx) = 1 + dy/dx

=> (2x - 2y) - (2x - 2y)dy/dx = 1 + dy/dx

=> dy/dx + (2x - 2y)dy/dx = 2x - 2y - 1

=> dy/dx(1 + (2x - 2y)) = 2x - 2y - 1

=> dy/dx = (2x - 2y - 1)/(2x - 2y + 1)(40 votes)- This is Sal's answer...

dy/dx = (2y - 2x + 1) / (2y - 2x -1)

If you change your answer like so...

dy/dx = (2x - 2y - 1) / (2x - 2y + 1)

dy/dx = -1(-2x + 2y +1) / -1(-2x + 2y -1)

dy/dx = (-2x + 2y +1) / (-2x + 2y -1)

dy/dx = (2y - 2x + 1) / (2y - 2x -1)

You can see that you and Sal both have the same answer.(116 votes)

- At2:24, did Sal accidentally forget the - when distributing (2x - 2y) into (1 - (dy/dx))?(34 votes)
- Nah

(2x-2y)(1-dy/dx)

(1-dy/dx)(2x-2y)

1(2x-2y)+(-dy/dx)(2x-2y)

(2x-2y)+(dy/dx)(-2x+2y)

(2x-2y)+(dy/dx)(2y-2x)(0 votes)

- 2:11, how come we can simplify the left side of the equation by simply distributing (2x - 2y)? Do we need to FOIL it? Or would it give the same results?(25 votes)
- Way, way, way back in polynomials, Sal teaches one common alternative to the FOIL method that is called distribution. In this case, we have (2x - 2y)(1 - dy/dx). The method is to split one of the binomials into its two terms and then multiply each term methodically by the two terms of the second binomial. So, as he says, multiply (2x - 2y) times 1 and (2x - 2y) times -1(dy/dx) to get (2x - 2y) + (2y - 2x)dy/dx = 1 + dy/dx

As you noticed, the result is the same, and it should be. It is just another way to methodically multiply binomials.(17 votes)

- Why is the d/dx multiplied by the constant ( in this case -1) equal to 0? Does it always equal 0 in every problem if it is a constant?(7 votes)
- There's a clue in the word 'constant'. d/dx multiplied by something is an operation to find this something's rate of change with respect to x. A constant, however, doesn't change at all, it stays constant! -1 constantly has the value -1. There are no changeable variables like x associated with it (like in, for example, -1x). So it has no rate of change. The rate of change of a constant always equals 0.(32 votes)

- At3:55, when Sal subtracts dy/dx from both sides, how does he end up with (2y-2x-1)dy/dx?(10 votes)
- (2y-2x) dy/dx - dy/dx = (2y-2x-1) dy/dx

It might help to imagine dy/dx as a single variable.

(2x - 2y)z - z = (2x - 2y - 1)z

If that doesn't help you may just want to expand and then re factor.

(2y-2x) dy/dx - dy/dx

2y dy/dx - 2x dy/dx - dy/dx Now factor out dy/dx

(2y - 2x - 1) dy/dx

Let me know if that didn't help.(24 votes)

- Hi everyone,

Quick Question - I'm stuck at3:40when Sal subtracts a dy/dx from the left hand side and gets -1dy/dx - how is this so?

In my mind dy/dx-dy/dx would somehow cancel eachother out but I think I've confused myself.(17 votes)- -dy/dx = -1 * dy/dx(5 votes)

- Shouldn't it be (2x-2y) - (2x-2y)(dy/dx) instead of (2x-2y)+(2x-2y)(dy/dx) at2:15(7 votes)
- He has switched the order of the variables, so that it is (2x-2y)+(2
**y**-2**x**)(dy/dx).

-(2x-2y) = 2y-2x(11 votes)

- At3:40i don't understand the process of subtracting the dy/dx(7 votes)
- The notation: dy/dx literally stands for the derivative of the function. Since it is an unknown you can treat it just as if it were a variable, which in a way, it is.

I hope this helps.(8 votes)

- At2:08Sal decides to distribute (2x - 2y) and at2:26he distributes as -> (2x - 2y) + (2x - 2y)(1 - dx/dy).

My concern revolves around the distribution, that it should be -> (2x - 2y) - (2x - 2y)(1 - dx/dy) because (2x - 2y) distributes itself into (1 - dx/dy).

Why has Sal put a + ?(3 votes)- Sal's work here was correct but used a little sleight of hand that's easy to overlook. For the second part of the expression, instead of writing what you were expecting

- (2x - 2y)(1 - dy/dx)

he wrote

+ (2y - 2x)(1 - dy/dx)

Notice that he reversed the order of the terms in the first parentheses: instead of 2x - 2y he wrote 2y - 2x. That reversal of order is equivalent to multiplying the expression by -1, and that allows him to change the minus sign to a +.(14 votes)

## Video transcript

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.