## The Chain of Calculus is Only as Strong as Its Weakest Link-- the Chain Rule

**You are about to learn the most important rule in differential calculus, and the second important in integral calculus.**

## Page Walkthrough

## Importance

Okay. I'm not that type of person that begs you to realize a theorem's importance. But here... things are different. This is, due to experience, the arguably most important principle in differential calculus. And it also plays a huge role in integral calculus as well.

*Go make yourself a cup of coffee (better without sugar), and make sure you understand every word on this page!*
## A guide to its usageNow, how do we use it? Think about it as a tool to break down functions. Instead of having to simplify a sometimes "unsimplifiable" function, we ignore parts of it, and try to identify a function that we are familiar with among the mess.
Beginners would be dazzled with possibilities: consider the function sqrt(1-x^2). Anybody with a little experience would immediately recognize the square root as the outer function, and 1-x^2 as the inner function. Now, to beginners, this is an obscure appliance because most would not recognize the pretty square root sign with the shape of a hut as a function. Now, practice is the solution. practice is the key. For after all, you have nothing to fear-- for you have found the weakest link in the seemingly unbreakable chain of calculus. |
## The chain ruleFirst consider a problem:
sin'(2x)=? If you are familiar with trigonometry, sin(2x) shall be immediately converted to 2sin(x)cos(x). Then we can easily crack up the problem using the product rule. Here, we present a much faster way: The derivative of f(g(x)) is equal to f'(g(x))*g'(x). Or, it is equal to the derivative of the outer function evaluated at the inner functions times the derivative of the inner function.MEMORIZE! So, for sin(2x). what is its derivative? cos(2x)*(2x)'. (sometimes, i get lazy, and represent the derivative of g(x)=2x by (2x)') This makes it 2cos(2x). As simple as that. You can apply the product rule as mentioned above and confirm this result. if you got it wrong, reconfirm. |

## Implicit Differentiation

I hate the name. It looks like we are doing something implicit here. Technically, that is the case.

We will extend our arm of power to implicit functions.

First a reminder: an explicit function is defined as y=something something x. all other functions are implicit functions.

Now, as I was saying, how do we calculate derivatives of implicit functions? I remembered that was an obstacle when i tried to study calculus. That is because, now we have to think of y, alongside of x, as something operable. we will illustrate that below.

Think of the function xy=1. what is the derivative? for the purposes here, we will not move y to the right. instead, we differentiate both sides.

Step one: use the product rule on the left side. xy'+y.

Step two: so we have xy'+y=0. now we isolate the variable y' as in algebra.

Step three: we have solved for the derivative in terms of both x and y. we can solve for y if we wish and substitute it back.

We will extend our arm of power to implicit functions.

First a reminder: an explicit function is defined as y=something something x. all other functions are implicit functions.

Now, as I was saying, how do we calculate derivatives of implicit functions? I remembered that was an obstacle when i tried to study calculus. That is because, now we have to think of y, alongside of x, as something operable. we will illustrate that below.

Think of the function xy=1. what is the derivative? for the purposes here, we will not move y to the right. instead, we differentiate both sides.

Step one: use the product rule on the left side. xy'+y.

__the derivative on the right side is just zero.__**(note that i have sued y' to denote the derivative of f(x). it will make this whole thing a lot simpler.)**Step two: so we have xy'+y=0. now we isolate the variable y' as in algebra.

Step three: we have solved for the derivative in terms of both x and y. we can solve for y if we wish and substitute it back.