The calculation of derivatives is common place in the fields of Calculus and Physics.  Derivatives are calculated to determine rates of change, velocities, accelerations, change in rates of growth and decay along with countless other applications in a wide range of fields.  By its definition, a derivative represents a change in the output of a function defined at some variable at some limit.

There are several rules of differentiation one can use to calculate a derivative, including but not limited to the Power, Product, Quotient and Chain rules.  Each rule is applicable to its own unique situation and multiple rules can be used along side one another when dealing with more complicated functions. The Chain Rule applies to situations where we have a function that is composed of two individual functions where one is a function of the other.  Each of these individual functions must have an associated derivative.

The Chain Rule formula states:

Let F(x) = j(k(x))

F’(x) = d/dx[j(k(x))] = j’(k(x))(k’(x))

To state this expression in words, the Chain Rule says that the derivative of function F (which consists of a function j defined with respect to function k which in tern is defined with respect to x) is the derivative of function j defined with respect to function k evaluated at x multiplied by the derivative of function k.

Chain Rule differentiation can be illustrated with an example.

Let’s say we want to find the derivative of F(x) = (5x2 – x + 10)1/2.

We will start by rewriting this function into the form of the Chain Rule.

F’(x) = d/dx[(5x2 – x + 10)1/2]( d/dx[5x2 – x + 10])

We will now apply the use of the Power rule to determine the derivative of each term and substitute the values back in to the Chain Rule formula.

d/dx[(5x2 – x + 10)1/2] = 1/2(5x2 – x + 10) -1/2

d/dx[5x2 – x + 10] = 10x –1

F’(x) = [½ (5x2 – x + 10) -1/2](10x –1)

Let’s see an additional example to illustrate how we can use the Power, Quotient and Chain rules altogether.

We want to find the derivative of F(x) = ((x + 2)/[(x2)(x – 1)])3

F’(x) = d/dx[((x + 2)/[(x2)(x – 1)])3]( d/dx[(x + 2)/[(x2)(x – 1)]])  (Chain rule form)

For the first term, we can differentiate using the power rule and substitute it back into the formula.

d/dx[((x + 2)/[(x2)(x – 1)])3] = 3((x + 2)/[(x2)(x – 1)])2

d/dx[((x + 2)/[(x2)(x – 1)])3] = 3((x + 2)/[(x3 – x2)])2

d/dx[((x + 2)/[(x2)(x – 1)])3] = 3(x + 2)2/(x3 – x2)2

d/dx[((x + 2)/[(x2)(x – 1)])3] = 3(x + 2)(x + 2)/(x3 – x2)(x3 – x2)

d/dx[((x + 2)/[(x2)(x – 1)])3] = 3(x2 + 4x + 4)/(x6 – 2x5 + x4)

For the second term, we will use the Power and Quotient rules to find the derivative and substitute it back into the Chain rule formula.

d/dx[(x + 2)/[(x2)(x – 1)]] =

[ ((x2)(x – 1)) d/dx[x + 2] – (x + 2) d/dx[(x2)(x – 1)] ] / [(x2)(x – 1)]2

d/dx[(x + 2)/[(x2)(x – 1)]] = [ (x3 – x2)(1) – (x + 2)(3x2 – 2x) ] / (x3 – x2)2

d/dx[(x + 2)/[(x2)(x – 1)]] = [ (x3 – x2) – (3x3 – 2x2 + 6x2 – 4x) ] / (x3 – x2)2

d/dx[(x + 2)/[(x2)(x – 1)]] = [–2x3 – x2 – 4x2 + 4x] / [(x3 – x2)(x3 – x2)]

d/dx[(x + 2)/[(x2)(x – 1)]] = (–2x3 – 5x2 + 4x) / (x6 – 2x5 + x4)

F’(x) = 3[(x2 + 4x + 4)/(x6 – 2x5 + x4)][ (–2x3 – 5x2 + 4x) / (x6 – 2x5 + x4) ]

F’(x) = 3[(x2 + 4x + 4)(–2x3 – 5x2 + 4x) ]/ (x6 – 2x5 + x4)2

Resource:

Stewart, James. (2001).  Calculus Concepts And Contexts (5th Ed.).  Pacific Grove, CA: Wadsworth Group.