Differentiability of Composite Functions

Composite Functions

Composite function is a function whose value is found from two given function. One function is applied to an independent variable and the second function is applied to the result. The domain of a composite function consists of those values of the independent variable for which the result yielded by the first function lies in the domain of the second function. An example of a composite function is g(f(x)). This can also be written as (g⁰ f)(x). Here one function is applied to the result of another function. In other words, the function g is applied to the result of the function f.

Properties of Composite Functions

The following properties of a composite function can easily be established:

1. Composite of functions is not commutative, that is, fog ≠ gof

2. Composite of functions is associative, that is, (fog)oh = fo(goh)

3. Composite of two bijective functions is also bijective.

4. If f and g are two bijective functions such that (gof) exists, then (gof)⁻¹ = f⁻¹og⁻¹ 

5. If f : X → Y is a bijective function, then f⁻¹ : X → Y is an inverse function of f. f⁻¹of = I\[_{x}\] and fof⁻¹ = I\[_{y}\].

6. When both f and g is even then, fog is an even function.

7. When both f and g is odd then, fog is an odd function.

8. If f is even and g is odd then, fog an even function.

9. If f is odd and g is even then, fog an even function.

Differentiability of a Function

A function f(x) is said to be differentiable at a point of its domain if it has finite derivative at that point. Thus f(x) is differentiable at x = a if \[\lim_{x\rightarrow a}\] \[\frac{f(x)-f(a)}{x-a}\] exists finitely.

Now,

\[\lim_{h\rightarrow a}\] \[\frac{f(a-h)-f(a)}{-h}\] = \[\lim_{h\rightarrow a}\] \[\frac{f(a+h)-f(a)}{h}\] 

⇒ f’(a-0) = f’(a+0)

Hence, left derivative = right derivative. Generally, derivative of f(x) at x = a is denoted by f’(a) . So,

f’(a) = \[\lim_{x\rightarrow a}\] \[\frac{f(x)-f(a)}{x-a}\]

Properties of Differentiable Functions

Properties of differentiable functions are as follows:

1. The sum, difference, product, quotient, (denominator

   
   0) and composite of two differentiable functions is always a differentiable function.

2. If a function is differentiable at some point, then it is necessarily continuous at that point, but its converse is not true. In other words, if a function is continuous at some point then it is not necessarily differentiable at that point.

Differentiability = continuity

Continuity ≠ differentiability

3. If a function is discontinuous at some point, then it is not differentiable at that point.

Solved Examples

Example 1:

If f(x) = \[\frac{1}{1-x}\], then find the derivate of the composite function f[f{f(x)}}].

Solution:

Given,

f(x) = \[\frac{1}{1-x}\]

Now, f{f(x)} can be evaluated by substituting \[\frac{1}{1-x}\] for x in the same equation. Therefore,

f{f(x)} = \[\frac{1}{1-\frac{1}{1-x}}\]

Simplifying,

f{f(x)} = \[\frac{1}{1-\frac{1}{1-x}}\]

f{f(x)} = \[\frac{1-x}{-x}\]

To find f[f{f(x)}}], substitute \[\frac{1}{1-x}\] for x in the above equation. Therefore,

f[f{f(x)}] = x

Simplifying,

f[f{f(x)}] = \[\frac{1-x-1}{1-x}\] x \[\frac{1-x}{-1}\]

f[f{f(x)}] = x

Differentiating f[f{f(x)}}],

\[\frac{d}{dx}\](x) = 1

Therefore, derivative of f[f{f(x)}}] is 1.

Example 2:

If g(x) is the inverse of an invertible function f(x) which is differentiable at x = c, then find the value of g’{f(c)}.

Solution:

Given:

 g(x) is the inverse of function f(x), therefore, it can be said that,

gof(x) = I(x) for all values of x,

Now,

gof(x) = I(x),∀x

⇒ gof(x) = x, ∀x = (gof)’(x) = I, ∀x

Using chain rule,

g’{f(x)}f’(x) = I, ∀x

⇒ g’{f(x)} = \[\frac{1}{f^{‘}(x)}\], ∀x

Substituting x = c in the above expression,

g’{f(c)} = \[\frac{1}{f^{‘}(c)}\] 

Example 3:

If f(x) = x(\[\sqrt{x}\] - \[\sqrt{x+1}\]), then

1. f(x) is continuous at x = 0 but not differentiable.

2. f(x) is differentiable when x = 0.

3. f(x) is not differentiable when x = 0.

4. None of the above

Solution:

For values of x greater than or equal to 0, the function f(x) will yield a value. Hence, f(x) is defined at x ≥ 0. Root of a negative number will give imaginary results. Therefore, f(x) is not defined for x < 0. Therefore, it can be said that the function f(x) is neither continuous nor differentiable at x = 0. 

Example 4:

f(x) = e\[^{x}\] ; x ≤ 0

If f(x) = l1 - xl ; x > 0, then

1. f(x) is differentiable at x = 0.

2. f(x) is continuous at x = 0.

3. f(x) is differentiable at x = 1.

4. f(x) is continuous at x = 1.

Solution:

The given function can be written as:

 = e\[^{x}\] ; x ≤ 0

f(x) = 1 - x ; 0 < x ≤ 1

= x - 1 ; x > 1

The right hand limit is evaluated as follows:

Rf’(0) = \[\lim_{h\rightarrow 0}\] \[\frac{f(0+h)-f(0)}{h}\] = \[\lim_{h\rightarrow 0}\] \[\frac{1-h-1}{h}\] = -1

The left hand limit can be evaluated as follows:

Lf’(0) = \[\lim_{h\rightarrow 0}\] \[\frac{f(0-h)-f(0)}{-h}\] = \[\lim_{h\rightarrow 0}\] \[\frac{e^{-h}-1}{-h}\] = 1

Since, left hand limit is not equal to the right hand limit, the function is not continuous at x = 0.

Similarly, it can be said that the given function is not differentiable at x = 1.

But it is continuous at x = 0,1.

Did you know

• If a function f(x) is differentiable at every point of the interval, it is said to be differentiable in an open interval (a, b).

• If a function f(x) is differentiable in (a, b) and has its right and left derivatives at a and b, respectively it is said to be differentiable in a closed interval [a, b].

• If the graph of a function y = f(x) is always smooth and has no breaks or corner, the function is derivable.

• The maximum or minima of a function at a point, makes the function differentiable at that point, and its derivative is zero.

• A composite function gof is defined if the range of f is a subset of the domain of g.

• Suppose f and g are two functions and h = fog. h will be differentiable if f and g is differentiable. If either f or g is not differentiable then h is not differentiable. Also, if both f and g are not differentiable, then h is not differentiable.