# Differentiability of Composite Functions

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].

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.

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., $fog\,\ne \,gof.$
• Composite of functions is associative i.e., $(fog)oh\,=\,fo(goh)$
• If $f:A\to B$ is bijection and $g:B\to A$  is inverse of f. Then $fog={{I}_{B}} \text \ and \ gof={{I}_{A}}.$where, ${{I}_{A}}$ and ${{I}_{B}}$ are identity functions on the sets A and B respectively.
• If $f:A\to B$ and $g:B\to C$ are two bijections, then $gof:A\to C$ is bijection and ${{(gof)}^{-1}}=({{f}^{-1}}o{{g}^{-1}}).$
• $fog\ne gof$ 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 xA.
• 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 be always 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.

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)

Limits Continuity and Differentiability

Differentiation

## Differentiability Of Composite Functions Examples

Example 1: Check whether the given function f (x) = |x| at x = 0 is differentiable.

Solution:

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).

Solution:

Let x < 0 ⇒ |x| = −x

F (x) = (d / dx) (x / [1 – x]) = 1 / (1−x)2

[f′(x)]x=0 = 1

Again

x>0

|x| = x

f (x) = (d / dx) (x / [1 + x]) = 1 (1 + x)2 ⇒ [f′(x)]x=0 = 1

f′(0)=1

Example 3: The set of all those points, where the function f (x) = x / [1+|x|] is differentiable, is _______________.

Solution:

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

Solution:

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 $f(x)=\left\{ \begin{matrix} {{e}^{x}}+ax, & x<0 \\ b{{(x-1)}^{2}}, & x\ge 0 \\ \end{matrix} \right.$ is differentiable at x = 0, then find the value of a and b.

Solution:

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

$\lim_{x\rightarrow 0^{-}}(e^{x}+ax)=\lim_{x\rightarrow 0^{+}}b(x-1)^{2} \\ e^0+a \times 0=b(0-1)^{2} \\ b=1$

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

$L{f}'(x)=R{f}'(x) \\ \frac{d}{dx}({{e}^{x}}+ax)=\frac{d}{dx}b{{(x-1)}^{2}} \\ {{e}^{x}}+a=2b(x-1) \text \ At \ x=0, {{e}^{0}}+a=-2b \\ a+1=-2b \\ a=-3 \\ (a,\,\,b)=(-3,\,\,1).$

Example 6: The function $f(x)=\left\{ \begin{matrix} {{e}^{2x}}-1 & , & x\le 0 \\ ax+\frac{b{{x}^{2}}}{2}-1 & , & x>0 \\ \end{matrix} \right.$ is continuous and differentiable for

A)            [.]

B)            a = 2, b = 4

C)            a = 2, any b

D)            Any a, b = 4

Solution:

Because f is continuous at x = 0, $f({{0}^{-}})=f({{0}^{+}})=f(0)=-1$

Also

$L{f}'(0)=R{f}'(0) \\ \underset{h\to 0}{\mathop{\lim }}\,\frac{f(0-h)-f(0)}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0+h)-f(0)}{h} \\ \underset{h\to 0}{\mathop{\lim }}\,\left( \frac{{{e}^{-2h}}-1+1}{-h} \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \frac{ah+\frac{b{{h}^{2}}}{2}-1+1}{h} \right) \\ \underset{h\to 0}{\mathop{\lim }}\,\left( \frac{-2{{e}^{-2h}}}{-1} \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( a+\frac{bh}{2} \right) \\ 2=a+0 \\ a=2,\,\,b$ any number.