Differentiability of Composite Functions

The composition of a function is an operation where two functions generate a new function. It is then not possible to differentiate them directly as we do with simple functions. This article explains the differentiability of composite functions along with solved examples.

Composite Function

Consider three sets, A, B and C, which are non-empty.

Let f: A → B and g: B → C be two functions. Then g of f: A → C. This function is called the composition of f and g. Read more.

Properties of the Composition of Functions

• f is even, and g is even – fog is an even function
• f is odd, and g is odd – fog is an odd function
• f is even, and g is odd – fog is an even function
• f is odd, and g is even – fog is an even function
• The composite of functions is not commutative, i.e., fog ≠ gof.
• The composite of functions is associative, i.e., (fog)oh = fo(goh)               
• If f : A → B is bijection and g : B → A  is the inverse of f. Then fog = I_B and gof = I_A, where I_A and I_B are identity functions on the sets A and B, respectively.
• If f : A → B and g : B → C are two bijections, then gof : A → C is a bijection and (gof)^-1 = f^-1og^-1.
• fog ≠ gof but if, fog= gof then either f^-1 = g or g^-1 = f also, (fog)(x) = (gof)(x) = x.
• (gof)(x) is simply the g-image of f(x), where f(x) is f-image of elements x ∈ A.
• Function gof will exist only when the range of f is the subset of the domain of g.
• fog does not exist if the range of g is not a subset of the domain of f.
• fog and gof may not always be defined.
• If both f and g are one-one, then fog and gof are also one-one.
• If both f and g are onto, then gof is onto.

How Do You Know if a Function Is Differentiable?

• The function f (x) is said to be differentiable at point P if and only if a unique tangent exists at point P.
• A function is said to be differentiable in an interval (a, b) if it is differentiable at every point of (a, b).
• A function is said to be differentiable in an interval [a, b] if it is differentiable at every point of [a, b].

Composite Function Differentiation

Let g and h be two functions where y = g (u) and u = h (x). If the function is defined by y = g [h (x)] or g o h(x), then it is called a composite function. So, if g (x) and h (x) are 2 differentiable functions, then fog is also differentiable, and hence, (fog)’(x) = f’(g(x)).g’(x)

Let y be a differential function of u, and u is a differential function of x, then

Let y = g(u) and u = f(x)

Let Δx be an increment in x, and Δ u and Δy be the corresponding increments in u and y, respectively.

Δy = [g(u + Δu)] – [g(u)] and Δu = [f(x + Δx) – f(x)] [Δy/(Δu)] = [(g (u + Δu)]/[(Δu)] and [Δu/(Δx)] = [(f(x + Δx) – f(x))]/[Δx] [Δy/(Δx)] = [Δy]/[(Δu).Δu/(Δx)]

On applying the limits,

⇒ [dy/dx] = [dy/du] × [du/dx]

= [d/du] g(u) × [d/dx] f(x)

Also, Read

Limits Continuity and Differentiability

Differentiation

Differentiability of Composite Functions Examples

Example 1: 

Find dy/dx if y = √(x² + 1).
Solution:

Example 2:  

Example 3: Show that function f (x) = (x² − 1) |x² − 3x + 2|+ cos (|x|) is not differentiable at points 0, 1 and 2.

Solution:

Since function |x| is not differentiable at x = 0,

|x² − 3x + 2| = | (x − 1) ( x − 2) |

Hence, it is not differentiable at x = 1 and 2

Now, f (x) = (x² − 1) |x² − 3x + 2|+ cos (|x|) is not differentiable at x = 2

For 1 < x < 2, f (x) = −(x² −1) (x² − 3x + 2) + cosx

For 2 < x < 3, f (x) = +(x² −1) (x² − 3x + 2) + cosx

L f′(x) = − (x² −1) (2x − 3) − 2x (x² − 3x + 2) − sinx

L f′(2) = − 3 − sin(2)

R f′(x) = (x² −1) (2x − 3) + 2x (x² − 3x + 2) − sinx

R f′(2) = (4 − 1) (4 − 3) + 0 − sin(2) = 3 − sin(2)

Hence, Lf′(2) ≠ Rf′(2).

Example 4: If

Solution:

Given f (x) is differentiable at x = 0. Hence, f (x) will be continuous at x = 0. 

But f (x) is differentiable at x = 0, then     

Video Lessons

Composite Functions

Composite and Periodic Functions