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 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 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., foggof.fog\,\ne \,gof.
  • Composite of functions is associative i.e., (fog)oh=fo(goh)(fog)oh\,=\,fo(goh)               
  • If f:ABf:A\to B is bijection and g:BAg:B\to A  is inverse of f. Then fog=IB and gof=IA.fog={{I}_{B}} \text \ and \ gof={{I}_{A}}.where, IA{{I}_{A}} and IB{{I}_{B}} are identity functions on the sets A and B respectively.
  • If f:ABf:A\to B and g:BCg:B\to C are two bijections, then gof:ACgof:A\to C is bijection and (gof)1=(f1og1).{{(gof)}^{-1}}=({{f}^{-1}}o{{g}^{-1}}).
  • foggoffog\ne gof but if, fog=goffog=gof then either f1=g{{f}^{-1}}=g or g1=f{{g}^{-1}}=f also, (fog)(x)=(gof)(x)=(x).(fog)\,(x)=(gof)\,(x)=(x).
  • gof(x)gof(x) is simply the g-image of f(x), where f(x) is f-image of elements xA.
  • Function gofgof will exist only when the range of f is the subset of the domain of g.
  • fogfog does not exist if the range of g is not a subset of the domain of f.
  • fogfog and gofgof may not be always defined.
  • If both f and g are one-one, then fogfog and gofgof are also one-one.
  • If both f and g are onto, then gofgof 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 the 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

[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)

Also Read

Limits Continuity and Differentiability

Differentiation

Differentiability Of Composite Functions Examples

Example 1: Find dy/dx if y = √(x2 + 1).
Solution:
dydx=12(x2+1)1/2×ddx(x2+1)\frac{dy}{dx} = \frac{1}{2}(x^2+1)^{-1/2} \times \frac{d}{dx}(x^2+1),
dydx=12(x2+1)1/2×(2x)\frac{dy}{dx} = \frac{1}{2}(x^2+1)^{-1/2} \times (2x),
dydx=x(x2+1)1/2\frac{dy}{dx} = x(x^2+1)^{-1/2}

Example 2: Find the derivative of 1cosx3\frac{1}{cosx^3}
Solution:
Given function is in the form f(g(h(x))). where f(x) = 1/x, g(x) = cos x and h(x) = x3
Let y = 1cosx3\frac{1}{cosx^3}.
Now, dydx=1(cosx3)2×ddx(cosx3)×ddx(x3)\frac{dy}{dx} = \frac{-1}{(cosx^3)^2} \times \frac{d}{dx}(cosx^3) \times \frac{d}{dx}(x^3) = (1(cosx3)2×[sin(x3)]×[3x2)(\frac{-1}{(cosx^3)^2} \times [-sin(x^3)] \times [3x^2) = (3x2sin(x3)(cosx3)2(\frac{3x^2sin(x^3)}{(cosx^3)^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, 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 f(x)={e2x+ax,x<0b(x1)2,x0f(x)=\left\{ \begin{matrix} {{e}^{2x}}+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. 

limx0(e2x+ax)=limx0+b(x1)2e0+a×0=b(01)2b=1\lim_{x\rightarrow 0^{-}}(e^{2x}+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     

Lf(x)=Rf(x)ddx(e2x+ax)=ddxb(x1)22e2x+a=2b(x1) At x=0,2e0+a=2ba+2=2ba=4(a,b)=(4,1).L{f}'(x)=R{f}'(x) \\ \frac{d}{dx}({{e}^{2x}}+ax)=\frac{d}{dx}b{{(x-1)}^{2}} \\ {2{e}^{2x}}+a=2b(x-1) \text \ At \ x=0, {2{e}^{0}}+a=-2b \\ a+2=-2b \\ a=-4 \\ (a,\,\,b)=(-4,\,\,1).

 

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