Last updated on July 15th, 2025
In calculus, the derivative of a composite function is determined using the chain rule, which allows us to understand how the composition of two functions changes in response to a slight change in the input variable. This concept is vital in various applications, such as physics and engineering, where functions of functions frequently occur. We will now explore the derivative of composite functions in detail.
To understand the derivative of a composite function, we consider two functions, f(x) and g(x), where f(g(x)) is the composite function. The derivative is typically represented using the chain rule: d/dx [f(g(x))] = f'(g(x)) · g'(x). This rule allows us to differentiate complex functions by breaking them down into simpler parts. Key concepts include: - Composite Function: A function consisting of two functions, f and g, such that f(g(x)). - Chain Rule: A fundamental differentiation rule for composite functions. - Differentiability: The property that indicates whether a function has a derivative at each point in its domain.
The formula for the derivative of a composite function is given by the chain rule: d/dx [f(g(x))] = f'(g(x)) · g'(x). This formula applies whenever both f and g are differentiable functions.
The derivative of a composite function can be proven using the chain rule. We demonstrate this using different methods: - By First Principle - Using the Chain Rule - Using the Product Rule We will now derive the differentiation of a composite function using these methods: By First Principle To prove using the first principle, consider f(g(x)). Its derivative by limit definition is: (f(g(x)))' = limₕ→₀ [f(g(x + h)) - f(g(x))]/h. Using the chain rule, we express this as: (f(g(x)))' = f'(g(x)) · g'(x). This result follows from recognizing the limits and applying the chain rule. Using Chain Rule The chain rule directly gives us the derivative: d/dx [f(g(x))] = f'(g(x)) · g'(x). This is the simplest and most direct method to differentiate composite functions. Using Product Rule Although not typically used for composite functions, the product rule can sometimes assist in understanding derivative relations within parts of the composite function. We consider: Let u = f(g(x)) and v = g(x), then: d/dx [u·v] = u'·v + u·v'. While this is less common for composite functions, it highlights the interrelation within the function parts.
When a composite function is differentiated several times, the resulting derivatives are called higher-order derivatives. These derivatives can be complex, as they require repeated application of the chain rule. For instance, the second derivative of a composite function involves: d²/dx² [f(g(x))] = [f''(g(x)) · (g'(x))²] + [f'(g(x)) · g''(x)]. Higher-order derivatives provide deeper insights into the behavior of composite functions.
Certain points or conditions might make the derivative of a composite function undefined or unique: - If g(x) is constant, the derivative f(g(x)) becomes zero. - If f is not differentiable at g(x), the derivative does not exist.
Differentiating composite functions can be tricky. Here are common mistakes and how to avoid them:
Calculate the derivative of (e^(sin x)).
Let f(x) = e^(sin x). Using the chain rule, we find: f'(x) = e^(sin x) · cos x. Thus, the derivative of e^(sin x) is e^(sin x) · cos x.
We identified the outer function as e^u and inner function as sin x. Applying the chain rule, we differentiated the outer function and multiplied it by the derivative of the inner function.
A water reservoir's volume is modeled by V(t) = ln(cos(t)). Find the rate of change of volume when t = π/3.
We have V(t) = ln(cos(t)). Differentiate using the chain rule: dV/dt = -tan(t). Substitute t = π/3: dV/dt = -tan(π/3) = -√3. Therefore, the rate of change of volume at t = π/3 is -√3.
We applied the chain rule to differentiate ln(cos(t)). Substituting t = π/3 gives the rate of change at that moment.
Find the second derivative of y = sin(x²).
First, find the first derivative: dy/dx = 2x cos(x²). Now, find the second derivative: d²y/dx² = 2 cos(x²) - 4x² sin(x²). Thus, the second derivative is 2 cos(x²) - 4x² sin(x²).
We first applied the chain rule to find dy/dx, then differentiated again, applying product and chain rules, to find the second derivative.
Prove: d/dx (cos³(x)) = -3cos²(x)sin(x).
Let y = cos³(x). Using the chain rule: dy/dx = 3cos²(x) · (-sin(x)). Thus, d/dx (cos³(x)) = -3cos²(x)sin(x).
We used the chain rule to differentiate, treating cos(x) as the inner function and its cube as the outer function, simplifying to the final result.
Solve: d/dx (ln(x^3 + 1)).
Differentiate using the chain rule: d/dx (ln(x^3 + 1)) = 3x²/(x^3 + 1). Thus, the derivative is 3x²/(x^3 + 1).
We applied the chain rule, treating x^3 + 1 as the inner function and ln as the outer function, then simplified.
Composite Function: A composition of two functions, represented as f(g(x)). Chain Rule: A rule used to differentiate composite functions. Differentiability: A function's ability to have a derivative at each point in its domain. Higher-Order Derivatives: Derivatives taken beyond the first derivative, often requiring repeated application of differentiation rules. Inner Function: The function g(x) within the composite function f(g(x)), which is differentiated first using the chain rule. ```
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