The function f (x) is said to be differentiable at point P, if and only if, a unique tangent exists at the point P. This article explains differentiability of composite functions.
In another way, f (x) is differentiable at point P, if and only if, P is not a corner point for the curve.
- 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].
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.
Properties of the composition of function
- f is even, and g is even – fog is even function
- f is odd, and g is odd – fog is an odd function
- f is even, and g is odd – fog is even function
- f is odd, and g is even – fog is even function
- Composite of functions is not commutative i.e.,
- Composite of functions is associative i.e.,
is bijection and is inverse of f. Then where, and are identity functions on the sets A and B respectively.
and are two bijections, then is bijection and but if, then either or also, is simply the g-image of f(x), where f(x) is f-image of elements x∈A.
will exist only when the range of f is the subset of the domain of g. does not exist if the range of g is not a subset of the domain of f. and may not be always defined.
- If both f and g are one-one, then
and are also one-one.
- If both f and g are onto, then
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[dy/dx] = [dy/(du)] × [du/(dx)]
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.[y + Δ y] = [g(u + Δu)] and [u + Δu] = [f(x + Δx)]
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)
Differentiability Of Composite Functions Examples
Example 1: Check whether the given function f (x) = |x| at x = 0 is differentiable.
Since this function is continuous at x=0
Now for differentiability,
f (x) = |x| = |0| = 0 and f (0+h) = f (h) = |h|
∴limh→0−[f (0+h) − f(0)] / [h] = limh→0− |h| / h = −1 and
limh→0+[f (0+h) − f(0)] / h = limh→0+|h| / h = 1
Therefore, it is continuous and non-differentiable.
Example 2: If f (x) = x / [1 + |x|] for x ∈ R, then find f′(0).
Let x < 0 ⇒ |x| = −x
F (x) = (d / dx) (x / [1 – x]) = 1 / (1−x)2[f′(x)]x=0 = 1
|x| = x
f (x) = (d / dx) (x / [1 + x]) = 1 (1 + x)2 ⇒ [f′(x)]x=0 = 1
Example 3: The set of all those points, where the function f (x) = x / [1+|x|] is differentiable, is _______________.
Let h(x) = x , x ∈ (−∞,∞); g(x) = 1 + |x|, x ∈ (−∞,∞)
Here “h” is differentiable in (−∞,∞) but |x| is not differentiable at x=0.
Therefore, “g” is differentiable in (−∞,0)∪(0,∞) and g(x)≠0,x∈ (−∞,∞), therefore
f (x) = h(x) / g(x) = x / [1+|x|]
It is differentiable in (−∞, 0) ∪ (0, ∞) for x = 0
limh→0 [f (h) − f (0)] / [ h − 0] = limh→0 [h / 1 + |h|] − 0 / h = limh→01 / [1 + |h|] = 1
Therefore, f is differentiable at x = 0, so f is differentiable in (−∞, ∞).
Example 4: The function f (x) = (x2 − 1) |x2 − 3x + 2|+ cos (|x|) is not differentiable at
A) 1 B) 0 C) 1 D) 2
Since function |x| is not differentiable at x = 0
∴|x2 − 3x + 2| = | (x − 1) ( x − 2) |
Hence, 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 − sin2
R f′(x) = (x2 −1) (2x − 3) + 2x (x2 − 3x + 2) − sinx
R f′(2) = (4 − 1) (4 − 3) + 0 − sin2 = 3 − sin2
Hence, Lf′(2) ≠ Rf′(2).
Example 5: If
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
Example 6: The function
B) a = 2, b = 4
C) a = 2, any b
D) Any a, b = 4
Because f is continuous at x = 0,