Implicit Differentiation Explained!

Added: 2026-06-04

[00:00:00]Today, let us understand a very important calculus topic called implicit differentiation.

[00:00:06]Normally in calculus, we write equations where Y is already isolated, like Y equals some function of X. These type of functions where Y is separated from X are called explicit functions. But sometimes equations contain both X and Y mixed together, and separating becomes difficult or impossible.

[00:00:29]In such cases, we use implicit differentiation. Let us start with a simple example. Suppose we have X squared plus Y squared equals 25. Our goal is to find dy over dx. We begin by differentiating both sides with respect to X. The derivative of X squared is 2X.

[00:00:50]Now comes the important part. The derivative of Y squared is not just 2Y.

[00:00:56]Since Y depends on X, we must multiply 2Y by dy over dx using the chain rule.

[00:01:04]The derivative of 25 is zero because constants always differentiate to zero.

[00:01:10]Now our goal is to isolate dy over dx.

[00:01:14]First, move the 2X term to the other side. This gives 2Y * dy over dx equals minus 2X.

[00:01:23]Next, divide both sides by 2Y.

[00:01:26]The twos cancel, and we finally get dy over dx equals minus X over Y.

[00:01:33]This is the derivative of the equation.

[00:01:36]Now let us try a slightly harder example involving the product rule. Suppose we have 3XY + Y squared equals 10.

[00:01:45]Again, we differentiate both sides with respect to X.

[00:01:49]The term 3XY contains multiplication of two variables. So we must use the product rule.

[00:01:57]The product rule says differentiate the first term and keep the second same.

[00:02:02]Then keep the first same and differentiate the second.

[00:02:06]So, differentiating 3x gives three while y stays the same.

[00:02:12]Then we keep 3x same and differentiate y which gives dy over dx.

[00:02:18]Next, differentiate y squared which becomes 2y multiplied by dy over dx.

[00:02:25]Finally, the derivative of 10 is zero.

[00:02:28]Next, move the term without dy over dx to the other side.

[00:02:33]Now, factor out dy over dx from the left side. This gives dy over dx multiplied by 3x + 2y = -3y.

[00:02:43]Finally, divide both sides by 3x + 2y.

[00:02:47]So, the answer becomes -3y / 3x + 2y.

[00:02:52]Nice.

[00:02:54]Let us now try an example involving trigonometric functions. Suppose sin of xy equals some constant c.

[00:03:02]Our goal again is to find dy over dx. We use the chain rule. So, differentiating sin of xy gives cosine of xy multiplied by the derivative of xy.

[00:03:16]Next, we will differentiate xy using the product rule. The derivative of x is one keeping y same gives y. Then keep x same and differentiate y which gives x multiplied by dy over dx. The derivative of c is zero. That leaves y + x multiplied by dy over dx = 0.

[00:03:40]Move y to the other side to get x multiplied by dy over dx = -y.

[00:03:48]Finally, divide by x. So, the derivative becomes minus y divided by x. I wasn't expecting this result.

[00:03:56]Sine and cosine simply disappeared. Now let us see another useful idea.

[00:04:02]Sometimes equations become easier if we simplify them before differentiating.

[00:04:07]Suppose we have 4 equals square root of x squared plus y squared. At first glance, differentiating this directly looks messy because of the square root.

[00:04:18]So instead, square both sides first.

[00:04:21]Then we get 16 equals x squared plus y squared.

[00:04:26]Now differentiation becomes much easier as it is similar to the first example we solved. The answer is simply minus x divided by y and that's it.

[00:04:37]If you enjoyed this video, please don't forget to like, share, and subscribe to our channel.

[00:04:44]So good.

[00:04:51]>> [music]