Chain Rule Blog 4

Hi class! Today we will be discussing the chain rule. 

The chain rule is used when working with compositions of functions and is therefore necessary for taking the derivative of any moderately complicated expression. 

The chain rule is defined as:

dz/dx = dz/dy (dy/dx)   or  f(g(x))' = f '(g(x)) (g'(x))

I understand that this definition may seem quite intimidating and intricate, but don't worry! As we work through it you will see how simple it really is!

To start off, let's say we have a function:

f(x) = (sinx)²

Our goal here isto find the derivative of the above function, or

f ’(x) = df/dx

What do you think we will do in order to find its derivative?

We’ll use the chain rule? You know, that crazy formula from the beginning of the lesson…

Maybe it will seem a bit less scary if I explain it in words. All that formula says is that in order to find the derivative of a composition of functions, you take the derivative of the outer function with respect to the inner function and multiply it by the derivative of the inner function with respect to x.

If we look back to our example of f(x) = (sinx)² , we can see that our inner function, our g(x),  is sinx.

So,   g(x) = sinx

We can also see that our outer function, our f(x), is x²

So,  f(x) = x²

From this, we can find our derivative, or f’(x) = df/dx

What we find is that

f’(x) = df/dx = 2sinx (cosx)

This is because the derivative of our outer function, x², with respect to our inner function, sinx, is 2sinx, or

f’(g(x)) =  the derivative of (sinx)² = 2sinx

and the derivative of our inner function, sinx, with respect to x is cosx, or

g’(x) = the derivative of sinx = cosx

So,

df/dx = d[(sinx)²]/ d(sinx) [d(sinx)/dx]

because d(sinx) cancels out on both sides, leaving us with

f’(x) = d[(sinx²)]/dx = 2sinx (cosx)

I hope that I was able to explain the concept of the chain rule to you all! If you have any further questions, feel free to ask away, as you will be using this concept many times in the near future!