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## AP®︎ Calculus BC (2017 edition)

### Course: AP®︎ Calculus BC (2017 edition)>Unit 4

Lesson 3: Implicit differentiation

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

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.
• At , did Sal accidentally forget the - when distributing (2x - 2y) into (1 - (dy/dx))? • , 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? • 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.
• 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? •  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.
• Hi everyone,
Quick Question - I'm stuck at when 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. • At , when Sal subtracts dy/dx from both sides, how does he end up with (2y-2x-1)dy/dx? • (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.
• Shouldn't it be (2x-2y) - (2x-2y)(dy/dx) instead of (2x-2y)+(2x-2y)(dy/dx) at • At i don't understand the process of subtracting the dy/dx • At Sal decides to distribute (2x - 2y) and at he 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 + ? • 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 +.