Understanding Differentiability in Composite Functions

The differentiability of composite functions is a foundational concept in differential calculus, essential for analyzing the behavior of nested or chained functions. To determine whether a composite function is differentiable and to compute its derivative, one must consider both the differentiability and continuity of the constituent functions involved in the composition.

Mathematical Statement of Differentiability for Composite Functions

Let $f : I \to \mathbb{R}$ and $g : J \to \mathbb{R}$ be real-valued functions with $g(I) \subseteq J$. The composite function is defined as $(f \circ g)(x) = f(g(x))$ for all $x \in I$.

If $g$ is differentiable at $x = a \in I$ and $f$ is differentiable at $y = g(a) \in J$, then the composite function $h(x) = f(g(x))$ is differentiable at $x=a$ and its derivative is given by

$\displaystyle h'(a) = \dfrac{d}{dx}\Big|_{x=a} f(g(x)) = f'(g(a))\cdot g'(a)$

Derivation of the Derivative Formula for Composite Functions

Let $h(x) = f(g(x))$. The derivative of $h$ at $x=a$ by first principles is defined as

$\displaystyle h'(a) = \lim_{x \to a} \dfrac{f(g(x)) - f(g(a))}{x - a}$

Introduce $y = g(x)$ and $y_0 = g(a)$. Observe that as $x \to a$, $g(x) \to g(a)$ due to the continuity of $g$ at $x=a$ (every function differentiable at a point is continuous at that point).

Rewrite the difference quotient as

$\displaystyle h'(a) = \lim_{x \to a} \frac{f(g(x)) - f(g(a))}{g(x) - g(a)} \cdot \frac{g(x) - g(a)}{x - a}$

This separates the limit into two components:

$\displaystyle h'(a) = \left( \lim_{x \to a} \frac{f(g(x)) - f(g(a))}{g(x) - g(a)} \right) \cdot \left( \lim_{x \to a} \frac{g(x) - g(a)}{x - a} \right)$

The first factor corresponds to $f'(g(a))$ as $g(x) \to g(a)$ when $x \to a$.

$\displaystyle \lim_{g(x) \to g(a)} \frac{f(g(x)) - f(g(a))}{g(x) - g(a)} = f'(g(a))$

The second factor is simply $g'(a)$.

$\displaystyle \lim_{x \to a} \frac{g(x) - g(a)}{x - a} = g'(a)$

Result: $\displaystyle h'(a) = f'(g(a)) \cdot g'(a)$

This formula is known as the chain rule for the differentiation of composite functions and forms the basis for calculating the derivative whenever a function is applied to the output of another function.

Conditions for Differentiability of Composite Functions

For the composite function $f(g(x))$ to be differentiable at $x = a$:

(1) The inner function $g$ must be differentiable at $x=a$.

(2) The outer function $f$ must be differentiable at $y = g(a)$.

(3) The function $g(x)$ must be such that $g(a)$ lies within the domain of $f$.

Stepwise Differentiation of Composite Functions Using the Chain Rule

Let $u = g(x)$ and $y = f(u)$. Then the composition is $y = f(g(x))$. Differentiating both sides with respect to $x$ gives

$\displaystyle \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}$

Explicitly, substitute $u = g(x)$ to obtain

$\displaystyle \frac{d}{dx}\left[f(g(x))\right] = f'(g(x)) \cdot g'(x)$

Explicit Calculation: Derivative of Threefold Composite Functions

Consider $h(x) = f(g(k(x)))$, where $k$ is differentiable at $x$, $g$ is differentiable at $k(x)$, and $f$ is differentiable at $g(k(x))$. The stepwise application of the chain rule is as follows.

Let $u = k(x)$, $v = g(u)$, and $y = f(v)$, so $y = f(g(k(x)))$.

Differentiate $y$ with respect to $x$ using the chain rule repeatedly:

$\displaystyle \frac{dy}{dx} = \frac{dy}{dv}\cdot\frac{dv}{du}\cdot\frac{du}{dx}$

Substitute back the variables to obtain

$\displaystyle \frac{d}{dx}[f(g(k(x)))] = f'(g(k(x)))\cdot g'(k(x))\cdot k'(x)$

Differentiability Of Composite Functions forms a critical subtopic for composite function analysis in examination settings.

Worked Example: Differentiation of a Composite Trigonometric Function

Given: $h(x) = \sin(x^2)$

Identify the inner and outer functions. Here, $f(u) = \sin u$ and $g(x) = x^2$.

Compute $f'(u) = \cos u$ and $g'(x) = 2x$.

Apply the chain rule:

$\displaystyle \frac{d}{dx} \sin(x^2) = f'(g(x))\cdot g'(x) = \cos (x^2) \cdot 2x$

Final result: $\displaystyle \frac{d}{dx} \sin(x^2) = 2x\cos(x^2)$

Worked Example: Derivative of a Composite Exponential Function

Given: $h(x) = e^{3x^2+1}$

Inner function: $g(x) = 3x^2+1$. Outer function: $f(u) = e^{u}$.

Compute $f'(u) = e^u$ and $g'(x) = 6x$.

By the chain rule,

$\displaystyle \frac{d}{dx} e^{3x^2+1} = f'(g(x)) \cdot g'(x) = e^{3x^2+1} \cdot 6x$

Final result: $\displaystyle \frac{d}{dx} e^{3x^2+1} = 6x e^{3x^2+1}$

Worked Example: Derivative of Three Composite Functions

Given: $f(x) = \ln \left(\sin(2x)\right)$

Let $u = 2x$, $v = \sin(u)$, and $y = \ln(v)$.

First, compute each derivative separately:

$\frac{dy}{dv} = \frac{1}{v}$

$\frac{dv}{du} = \cos u$

$\frac{du}{dx} = 2$

By the chain rule,

$\displaystyle \frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{du} \cdot \frac{du}{dx} = \frac{1}{\sin(2x)} \cdot \cos(2x) \cdot 2$

Simplify the expression:

$\displaystyle \frac{dy}{dx} = \frac{2\cos(2x)}{\sin(2x)}$

Final result: $\displaystyle \frac{d}{dx} \ln \left(\sin(2x)\right) = 2~\cot(2x)$

Continuity and Differentiability Relationship in Composite Functions

If $g$ is continuous at $x=a$ and $f$ is continuous at $g(a)$, then $f \circ g$ is continuous at $x=a$.

If, moreover, $g$ is differentiable at $x=a$ and $f$ is differentiable at $g(a)$, then $f \circ g$ is differentiable at $x=a$ and its derivative is computed using the chain rule as previously established.

Partial Derivative of Composite Functions Involving Several Variables

Consider $z = f(g(x, y))$, where $g$ is a differentiable function of two variables and $f$ is differentiable at $g(x, y)$. The partial derivatives of $z$ with respect to $x$ and $y$ are given by

$\displaystyle \frac{\partial z}{\partial x} = f'(g(x, y)) \cdot \frac{\partial g(x, y)}{\partial x}$

$\displaystyle \frac{\partial z}{\partial y} = f'(g(x, y)) \cdot \frac{\partial g(x, y)}{\partial y}$

These generalizations of the chain rule allow for differentiation of composite functions in higher-dimensional settings and form the basis for implicit differentiation and advanced analysis.

Comprehension of the Differential Calculus and composite differentiation supports advanced problem solving in calculus and related fields.