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# Worked example: Implicit differentiation

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.D (LO)
,
FUN‑3.D.1 (EK)
Implicit differentiation of (x-y)²=x+y-1. Created by Sal Khan.

## Want to join the conversation?

• At , did Sal accidentally forget the - when distributing (2x - 2y) into (1 - (dy/dx))?
• 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)
• 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.
• , 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.
• 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.
• -dy/dx = -1 * dy/dx
• Shouldn't it be (2x-2y) - (2x-2y)(dy/dx) instead of (2x-2y)+(2x-2y)(dy/dx) at
• He has switched the order of the variables, so that it is (2x-2y)+(2 y-2 x)(dy/dx).

-(2x-2y) = 2y-2x
• At i don't understand the process of subtracting the dy/dx
• 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.
• 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.
• 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.
• 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 + ?
(1 vote)
• 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 +.