Let’s study Composite Function Differentiation

A composite function is a combination of two different functions to produce a third function. Let us consider two functions u(x) and v(x). In this v(x) is acting first and then the u(x). It is written as u(v(x)) or u∘v(x).

In composite functions, the output of the inner function becomes the input of the outer function. Let us understand:

• In u[v(x)], v(x) is the input for the outside function u(x). It is also written as (u ∘ v)(x) and is read as ‘u of v of x’.
• In v[u(x)], u(x) is the input for the outside function v(x). It is also written as (v ∘ u)(x) and is read as ‘v of u of x’.

Derivatives of composite functions are calculated using the chain rule of differentiation. To determine the differential of the composite function, firstly we differentiate the inner function w.r.t the outer function and then differentiate the inner function w.r.t the variable. For example –

(f ∘ g )′ (x) = f ′ (g(x)).g ′ (x)

Another method to write is as follows:

Let h be a composite function having real values. It consists of two functions f and g. Such as h = f ∘ g. Let u = g(x) where du/dx and df/du exist, then 

dh/dx = df/du, where h = f(u) and u = g(x)

Simplifying it further, we get

dh/dx = df/du . du/dx

Alternate way to write the derivative of h(x) is:

h′(x) = f ′ (g(x)) .  g′(x)

Hence we can say that if f and g are functions having real values, then the derivative of (f ∘ g) is expressed as :

(f ∘ g) ′ = d/dx [f (g(x))] = f ′ [g(x)] . g ′ (x) = (f ′ ∘ g) . g ′ 

The differential of composite functions having one variable

Derivatives of composite functions consisting of only one variable are calculated using the simple chain rule. While using the chain rule, we use the BODMAS. This means that first whatever is there in the inner function has to be calculated and the output of the inner function is substituted in the outer function.

Let us consider a few examples to understand the concept.

Example 1: Calculate the derivative of the composite function m(x) = (x⁴ + 8)¹²

Solution: Now, let u = x⁴ + 8 = g(x), 

here m(x) can be written as m(x) = f(g(x)) = u¹².

So the derivative of m(x) is given by:

d(m(x))/dx = df/du × du/dx

⇒ m ‘ (x) = 12 u¹¹ × 4x³

= 12(x⁴ + 8)¹¹ × 4x³

= 48 x³ (x⁴ + 8)¹¹

Example 2: Derivative of composite function y = sin (cos (x³))

Solution: y’ = cos(cos (x³)). -sin (x³)). 3x²

= -3x² sin (x³). cos (cos (x³))

Example 3: Find dy/dx if y = √x² + 1. 

Solution: The outside function is √(x³ + 2 = (x³ + 2)¹/² which has derivative 1/2 (x³ + 2)⁻¹/² , 

The inside function is (x³ + 2) so that y = 1/2 (x³ + 2)⁻¹/² × 3x²

Alternatively, if u = x³ + 2, 

we have y = √u =(u) ¹/² 

So dy/dx =1/2 ( u)⁻¹/² × 3x² = 1/2 (x³ + 2)⁻¹/² ×3x².

Partial derivatives of composite functions having two variables

The differential of composite functions with many variables is calculated w.r.t one of the variables keeping the other variables constant at one time. Such derivatives are known as partial derivatives. We can calculate the partial differential of composite functions m = n(x,y) by using the chain rule method. First, we calculate the derivatives with x as the variable while keeping y as a constant, and then we keep x as constant and use y as a variable.

Example: Find the derivatives of the composite function f(m, n) = (m3n2 + ln m)2

Solution: First, we will find the differential of the composite function f(m, n) = (m³n² + ln m)² with respect to m and consider n as a constant.

∂[(m³n² + ln m)²]/∂m = 2 (m³n² + ln m) × ∂(m³n² + ln m)/∂m

= 2 (m³n² + ln m) × (3m²n² + 1/m)

= 2(3m²n² + 1/m)(m³n² + ln m)

Now, we will determine the n-derivative considering m as a constant.

∂[(m³n² + ln m)²]/∂n = 2 (m³n² + ln m) × ∂(m³n² + ln m)/∂n

= 2 (m³n² + ln m) × (2m³n)

= 4m³n (m³n² + ln m)

Conclusion

We have seen thus far that we can calculate the derivatives of composite functions with one or more variables using the chain rule method. Let us recap the same.

• The derivative of a composite function h = f(g(x)) where f(x) and g(x) are two differentiable functions. The differential of h w.r.t x is given as-

dh/dx = f ′ (g(x)). g′(x)

• The a-derivative of a composite function u = g(x(a), y(a)) can be calculated by using the chain rule and partial derivative of one variable with the other variable as constant and vice versa.

du/da = (∂g/∂x) . dx/da + (∂g/∂y) . (dy/da)

• Let h = (f ∘ g) and u = g(x) . Then the derivative of h w.r.t  x is written as

dh/dx = df/du . du/dx