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## Differential Calculus

### Unit 3: Lesson 4

Implicit derivative of (x²+y²)³=5x²y². Created by Sal Khan.

## Want to join the conversation?

• Why, early in the equation solving, was the 5 taken out of the derivative, , and then multiplied back in later, , instead of performing the derivative of 5x^2y^2 when taking the derivative of everything else? Although I understand the the ending result and the concept this small point is a bit confusing to me in relation to the rest of the problem and the prior videos. •  Novamwilliams2,

I'm not sure what to say besides it being a matter of taste. Sal could have chosen to do the product rule using 5x^2 and y^2 as his two functions that were being multiplied. In general I see people factor out any constants and then bring them back after all the calculus is done. I would not be surprised if Sal is not consistent with how he does this.
• How is the derivative of y^2 with respect to x : 2y (dy/dx)? • It's another chain rule thing, because it applies when you're taking the derivative of something, so y^2 becomes:
(2y^(2-1)) • (derivative of y with respect to x)
or:
2y • (dy/dx)

Similarly, y^1 in the same situation would go through the chain rule, but would cancel itself out via its exponent being zero:
(1y^(1-1)) • (derivative of y with respect to x)
or:
(1) • (dy/dx)
• Why does Sal keep distributing terms like this at ? I keep thinking he should be using FOIL, but it seems like he is treating binomials as if they are a single term? • He could simplify the end expression abit right? atleast divide it by 2.
Is there a spesific reason he didnt do that? • In calculus, unlike some earlier levels of math, it is not at all uncommon to leave an expression in something other than its simplest form. In fact, even what counts as "the simplest form" may be ambiguous.

So, in Calculus, you think more in terms of whether the form is accurate, informative, useful, or practical. As such, not fully simplifying the answer is not at all unusual.
• For the final answer... if you subbed in the x and y values for a given point to find the slope of the tangent line, wouldn't you get more than one answer? This isn't a function it doesn't pass the vertical line test. There are multiple values for y when given an x value. • That is exactly why EXPLICIT differentiation doesn't work and you need to use IMPLICIT differentiation because the gradient function depends on BOTH the x value and y value.

If you only have the x-value, then you have 4 possible answers in this case. However, to find the gradient of the point that Sal plotted, you first need to find its y-coordinate as well using the original function (or u will be given the y-coordinate if you are lucky).

The way you do this is solve the function for the y-value at the given x using a graphing or solving capabilities of calculators or computer programs. You can visually see that the y-coordinate of the point that Sal plottes is POSITIVE and the 2nd GREATEST value, thus, you can input both the x and y coordinates into the gradient function to find the gradient.

I really hope this makes a little sense.
• The applying chain rule to y^2 thing. Is there a reason? Does it always go that you have to use chain rule to get the derivative of something that has y? • Yes. The whole point of implicit differentiation is to differentiate an implicit equation, that is, an equation that is not explicitly solved for the dependent variable 𝑦. So whenever we come across a 𝑦 term when implicitly differentiating, we must assume that it is a function of 𝑥. So by assuming it is a function of 𝑥 (without knowing the function explicitly), we differentiate 𝑓(𝑦) as a composition of functions 𝑓 and 𝑦. In this case, 𝑓(𝑥) = 𝑥², so to differentiate 𝑓(𝑦), we note that 𝑦 itself is a function of 𝑥 and we have by chain rule:
[𝑓(𝑦)]' = 𝑓'(𝑦) • 𝑦' = 2𝑦 • 𝑦'
The key is to note that we are always assuming that 𝑦 is a function of 𝑥 when we implicitly differentiate without explicitly solving for that function. So whenever we find something "being done to" 𝑦, we basically have a function composition of 𝑦 and the function that describes whatever is being done to it. And of course, to differentiate a composition, you need chain rule!

I hope you understand the logic behind this! Many people just "memorize" the steps to implicit differentiation without understanding why 𝑦 terms are treated the way they are. Feel free to comment if you have questions!
• Would:

dy/dx = (6x*(x^2+y^2) - 10x*y^2) / ( (10*(x^2)y) - 6*y(x^2+y^2)^2 )

be equally valid? • Sal derives two things in the second bracket. X^2 with respect to x and Y^2 with respect to why. Can you even do that? same bracket derived with respect to different values?
(1 vote) • Are you not able to multiply 6x by (x^2y^2)^2 to get 6(x^3y^2)^2? • I'm a little confused where you saw (x^2y^2)^2. What I saw was 6x(x^2 + y^2)^2. That plus in there makes a huge difference. Otherwise still no, you have to perform exponents FIRST (PEMDAS). Which requires you take the exponent of (x^2y^2) first, giving you x^4y^4, then you may multiply by 6x, giving you 6x^5y^4.

Regardless I didn't see an (x^2y^2)^2 in the video, I saw (x^2 + y^2) ^2, which is an entirely different story since you must foil that (x^2 +y^2) first, not simply raising the terms to powers and giving the false result of (x^2 + y^2)^2 = x^4 + y^4; <-- that is incorrect since you didn't foil!! 