The Chain Rule is how we combine functions together and use those combinations to give us results. We can use the Chain Rule to Differentiate functions that we otherwise would not have the method to differentiate.

If *y* is a differentiable function of *u*, say y = f(u) and *u* is a differentiable function of *x*, say u = f(x) then *y* is a differentiable function of *x*.

dy

dx

=

dy

du

.

du

dx

In other words, we can get to the differentiable value of *x* from *y*, by going via *u*. *u* is connected to *y* and connected to *x*. Think of it like,

Yvonne knows Umar,

and Umar knows Xavier,

so Yvonne knows Xavier through Umar

One other point we can note that may help us understand why the Chain Rule works is by thinking of it like,

5

7

=

5

1

.

1

7

This can also be mathematically written as follows:

5

7

=

5

1

.

1

7

.

du

du

Think of the "du" as an irrelevance since, just like the 1, it will cancel out.

=

5

du

.

du

7

It still equals the same thing, however, mathematically it has use to us.

Now think of it like:

dy

dx

=

dy

du

.

du

dx

It may seem pointless to add in "du" as it can cancel out. However we will see this can be used to help us in differentiation.

The Chain Rule makes more sense once it is seen in an example.

In this example, we will use Leibniz Chain Rule to differentiate an equation that can also be done simply by multiplying out the brackets and then differentiating. The purpose of this example is to introduce you to Leibniz Chain Rule. We will use the Chain Rule when it is not possible to differentiate using standard methods.

Use the Leibniz Chain Rule to differentiate:

y = 2(2x + 3)^{2}

Substituting u = 2x + 3

We can differentiate by expanding the brackets but for the purpose of practicing the Leibniz Rule will do the substitution.

So first we substitute u = 2x + 3

y = 2(2x + 3)^{2}

Becomes:

y = 2(u)^{2}, y = 2u^{2}

So now we have an equation with y & u and u = 2x + 3, u & x

When we differentiate y = 2u^{2} and u = 2x + 3

We get = (2)2u^{2-1} = 4u & = 2

Recall Leibniz Chain Rule:

dy

dx

=

dy

du

.

du

dx

So:

Now think of it like:

dy

dx

= 4u.2 = 8u

Now, lastly, substitute *u = 2x + 3* back in.

dy

dx

= 8(2x + 3) = 16x + 24

If you like, try differentiating the original equation.

y = 2(2x + 3)^{2} = 2(4x^{2} + 12x + 9)

= 8x^{2} + 24x + 18

= 16x + 24

Follow the steps that use the Chain Rule to answer each of the two questions below.

If we have a function, g(x) that is differentiable at *x*, and* f* is a function differentiable at g(x),

Then the composite *f* of *g* is differentiable at *x* and therefore,

(f of g)ā(x) =**D**( f( g(x) ) ) = fā( g(x) )( gā(x) )

As in the Leibniz Chain Rule, the first example will be one to familiarize yourself with the Composition Chain Rule.

Use the Composite Chain Rule to differentiate:

y = 6

x^{2} + x

First, recall the Composite Chain Rule:

D(f(g(x))) = f'(g(x)).g'(x)

So we must identify an inner and outer function.

For,

y = 6

x^{2} + x

Try to see where there are two functions.

y = 6

x^{2} + x

We have the outer function,

y = 6

(something)

, and the inner function, x^{2} + x

We can write these as, f(x) = 6

x

, and g(x) = x^{2} + x

Now we calculate their derivatives.

f(x) = 6(x)^{½}

f'(x) = (½)6(x)^{½ - 1}

f'(x) = 3x^{-½}

=

3

x

g(x) = 2x + 1

Now sub these into:

D(f(g(x))) = f'(g(x)).g'(x)

=

3(2x + 1)

(x^{2} + x)

=

6x + 3

(x^{2} + x)

Follow the steps in the two worked examples below to see the use of the Product Rule and Chain Rule to find derivatives of functions.

The two examples below show the steps to differentiate functions with arctan using the Chain Rule.

The two examples below show the steps to differentiate functions with arctan using the Chain Rule.

Follow the steps in the two worked examples below to see the Chain Rule used with sinh, cosh, and ln.