# The Chain Rule

Contents

### Summary

• Chain rule lets us differentiate a function of a function i.e $f(g\left( x \right) )$
• Chain rule can be applied through two different equations: $\frac { d }{ dx } \left[ f(g\left( x \right) ) \right] \quad =\quad f^{ ' }\left( g\left( x \right) \right) \quad \times \quad g^{ ' }\left( x \right)$

AND $\frac { dy }{ dx } \quad =\quad \frac { dy }{ du } \quad \times \quad \frac { du }{ dx }$

#### What is the chain rule?

One of the rules of differentiation is the chain rule, which basically lets us differentiate a function of a function in other words differentiation of composite functions.

Let’s suppose we have a function y which is formed by two different functions f(x) and g(x) i.e $f(g\left( x \right) )$f(x) is said to be the outer function and g(x) is the inside function.

You will be able to understand this better when we go through an example.

#### Example #1

Let $y\quad =\quad { (2x\quad +\quad 3{ x }^{ 3 }) }^{ 8 }$

We can see that in this case the outer function f(x) is ${ x }^{ 8 }$  and the inside function g(x) is $2x\quad +\quad 3{ x }^{ 3 }$.

When we put g(x) inside the expression of f(x) , like $f{ (2x\quad +\quad 3{ x }^{ 3 }) }$  we get: $f(g\left( x \right) )\quad =\quad y\quad =\quad { (2x\quad +\quad 3{ x }^{ 3 }) }^{ 8 }$

#### Example #2 $y\quad =\quad { cos^{ 4 }(x) }$

As ${ cos^{ 4 }(x) }$  is actually ${ (cos(x)) }^{ 4 }$

Therefore: $f\left( x \right) \quad =\quad { x^{ 4 } }$ $g\left( x \right) \quad =\quad cos(x)$

Hence: $f(g\left( x \right) )\quad =\quad f(cos(x))$ $=\quad { cos^{ 4 }(x) }$

Now that we have understood what it means by a ”function of a function” or composite function, we can move on to the chain rule technique.

If we are able to differentiate function g with respect to x and we are able to differentiate function f with respect to g(x), then the composite function $f(g\left( x \right) )$  is differentiable and we can use the chain rule technique to find its derivative.

#### The chain rule formula is defined as: $\frac { d }{ dx } \left[ f(g\left( x \right) ) \right] \quad =\quad f^{ ' }\left( g\left( x \right) \right) \quad \times \quad g^{ ' }\left( x \right) \quad \rightarrow \quad eq\quad 1$

Where g'(x) is the derivative of the function g(x),
and $f^{ ' }\left( g\left( x \right) \right)$  is the derivative of the function $f(g\left( x \right) )$.

#### The following three steps explains how the formula is applied:

i) Differentiate the outside function f(x)
ii) Leave the inside function g(x) alone
iii) Multiply by the derivative of the inside function g(x)

#### There is also another way to apply the chain rule:

to find the derivative of $y\quad =\quad f(g\left( x \right) )$  we take g(x) and substitute it as u i.e:

Let u = g(x)

Therefore now y = f(u).

Hence, to find $\frac { dy }{ dx }$  we can apply chain rule formula $\frac { d }{ dx } \left[ f(g\left( x \right) ) \right] \quad =\quad f^{ ' }\left( g\left( x \right) \right) \quad \times \quad g^{ ' }\left( x \right)$  in a way that it becomes: $\frac { dy }{ dx } \quad =\quad \frac { dy }{ du } \quad \times \quad \frac { du }{ dx } \quad \rightarrow \quad eq\quad 2$

After we go through a few examples using both the formulas, you can choose which one you find easy to apply.

#### Example #3

Q. Find the derivative of $y\quad =\quad cos{ (x) }^{ 2 }$

We know f(x) = cos(x) $g\left( x \right) \quad =\quad { x^{ 2 } }$

Using equation 1 here: $\frac { d }{ dx } \left[ f(g\left( x \right) ) \right] \quad =\quad f^{ ' }\left( g\left( x \right) \right) \quad \times \quad g^{ ' }\left( x \right)$

i) f'(x) = -sin(x)

ii) $g\left( x \right) \quad =\quad { x^{ 2 } }$

iii) g'(x) = 2x

Put the above values in the equation: $\frac { d }{ dx } \left[ f(g\left( x \right) ) \right] \quad =\quad f^{ ' }\left( { x^{ 2 } } \right) \quad \times \quad g^{ ' }\left( x \right)$ $=\quad -sin({ x^{ 2 } })\quad \times \quad 2x$

Ans: $=\quad -2xsin({ x^{ 2 } })$

#### Example #4

Q. Now solve the same question using equation 2

As $y\quad =\quad cos{ (x) }^{ 2 }$

Let $u\quad =\quad x^{ 2 }$

Therefore:

y = cos u

Use equation 2: $\frac { dy }{ dx } \quad =\quad \frac { dy }{ du } \quad \times \quad \frac { du }{ dx }$

To find $\frac { dy }{ dx }$  we first find the following two: $\frac { dy }{ du } \quad =\quad -sin\quad u$  and $\frac { du }{ dx } \quad =\quad 2x$

Substitute the values in the equation: $\frac { dy }{ dx } \quad =\quad -sin\quad u\quad \times \quad 2x$ $=\quad -2xsin\quad u$   substitute $u\quad =\quad x^{ 2 }$ $=\quad -2xsin(x^{ 2 })\quad$

Hence, this shows we reach the same answer using both the equations, so now you can decide which one is easier for you.