Derivative of f(g(x))
2026-02-28 06:18 Diff
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Last updated on August 5, 2025

We use the derivative of a composite function, f(g(x)), to understand how the composition of functions changes in response to a slight change in x. Derivatives assist in various real-life applications, such as calculating the rate of change in different scenarios. We will now discuss the derivative of f(g(x)) in detail.

What is the Derivative of f(g(x))?

We now explore the derivative of the composite function f(g(x)). It is commonly represented as d/dx [f(g(x))] or [f(g(x))]', and its value depends on the derivatives of both f and g. The function f(g(x)) is differentiable within its domain if both f(x) and g(x) are differentiable. The key concepts are mentioned below: Composite Function: A function composed of two functions, f and g, where f(g(x)) is the composition. Chain Rule: A rule for differentiating composite functions. Derivative: Represents the rate of change of a function concerning its variable.

Derivative of f(g(x)) Formula

The derivative of f(g(x)) is denoted as d/dx [f(g(x))] or [f(g(x))]'. The formula used to differentiate f(g(x)) is: d/dx [f(g(x))] = f'(g(x)) · g'(x) This formula applies to all x where both f and g are differentiable.

Proofs of the Derivative of f(g(x))

We can derive the derivative of f(g(x)) using various proofs. To show this, we will use differentiation rules. Some methods for proving this include: By Chain Rule By Substitution We will now demonstrate that the differentiation of f(g(x)) results in f'(g(x)) · g'(x) using these methods: Using Chain Rule To prove the differentiation of f(g(x)) using the chain rule: Consider f(x) and g(x), where the composite function is f(g(x)). By the chain rule: d/dx [f(g(x))] = f'(g(x)) · g'(x) This formula states that to differentiate a composite function, multiply the derivative of the outer function evaluated at the inner function by the derivative of the inner function. By Substitution Another method involves substitution: Let h(x) = g(x), so f(g(x)) = f(h(x)). The derivative is: d/dx [f(h(x))] = f'(h(x)) · h'(x) Substituting h(x) back for g(x), d/dx [f(g(x))] = f'(g(x)) · g'(x) Hence, proved.

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Higher-Order Derivatives of f(g(x))

When a function is differentiated multiple times, the derivatives obtained are higher-order derivatives. To understand them, consider a scenario where the speed (first derivative) and the rate of change of speed (second derivative) change. Higher-order derivatives provide insights into the behavior of functions like f(g(x)). For the first derivative, we write f′(x), indicating the function's rate of change or slope. The second derivative, f′′(x), is derived from the first derivative, indicating the rate of change of the rate of change. This pattern continues for higher-order derivatives. For the nth derivative of f(g(x)), we generally use fⁿ(x) to represent the nth derivative, which tells us about the change in the rate of change.

Special Cases:

When g(x) has points where its derivative is undefined, the derivative of f(g(x)) is also undefined at those points. If g(x) is a constant, the derivative of f(g(x)) depends solely on the derivative of the constant function.

Common Mistakes and How to Avoid Them in Derivatives of f(g(x))

Students frequently make mistakes when differentiating f(g(x)). These mistakes can be resolved by understanding the correct methods. Here are a few common mistakes and ways to solve them:

Problem 1

Calculate the derivative of f(g(x)) where f(x) = x² and g(x) = sin(x).

Okay, lets begin

Here, we have f(g(x)) = (sin(x))². Using the chain rule: d/dx [f(g(x))] = f'(g(x)) · g'(x) f'(x) = 2x and g'(x) = cos(x), so: d/dx [(sin(x))²] = 2(sin(x)) · cos(x) Therefore, the derivative of (sin(x))² is 2sin(x)cos(x).

Explanation

We find the derivative of the given function by applying the chain rule. The first step is finding the derivative of each function and then combining them using the chain rule to get the final result.

Well explained 👍

Problem 2

A bridge's height from the ground is represented by the function h(x) = e^(tan(x)), where x is the distance from the start. If x = π/4, calculate the rate of change of the bridge's height.

Okay, lets begin

We have h(x) = e^(tan(x)). Differentiating h(x): dh/dx = e^(tan(x)) · sec²(x) Given x = π/4, dh/dx = e^(tan(π/4)) · sec²(π/4) = e^1 · 2 (since tan(π/4) = 1 and sec²(π/4) = 2) = 2e Hence, the rate of change of the bridge's height at x = π/4 is 2e.

Explanation

We find the rate of change of the bridge's height by differentiating the given function using the chain rule. We substitute x = π/4 to calculate the rate of change at that specific point.

Well explained 👍

Problem 3

Derive the second derivative of the function y = ln(cos(x)).

Okay, lets begin

The first step is to find the first derivative: dy/dx = -tan(x)...(1) Now we differentiate equation (1) to get the second derivative: d²y/dx² = -sec²(x) Therefore, the second derivative of the function y = ln(cos(x)) is -sec²(x).

Explanation

We use a step-by-step process, where we start with the first derivative. We then differentiate the result to find the second derivative, applying differentiation rules as necessary.

Well explained 👍

Problem 4

Prove: d/dx [cos²(x)] = -2cos(x)sin(x).

Okay, lets begin

Let’s start using the chain rule: Consider y = cos²(x) = [cos(x)]² To differentiate, we use the chain rule: dy/dx = 2cos(x) · (-sin(x)) = -2cos(x)sin(x) Hence proved.

Explanation

In this step-by-step process, we used the chain rule to differentiate the equation. We replaced cos(x) with its derivative and simplified the result.

Well explained 👍

Problem 5

Solve: d/dx [e^(x²)].

Okay, lets begin

To differentiate the function, we use the chain rule: d/dx [e^(x²)] = e^(x²) · 2x Therefore, d/dx [e^(x²)] = 2xe^(x²).

Explanation

In this process, we differentiate the given function using the chain rule. We apply the derivative of the exponent and multiply it by the derivative of x² to obtain the final result.

Well explained 👍

FAQs on the Derivative of f(g(x))

1.Find the derivative of f(g(x)).

Using the chain rule for the composite function f(g(x)), we get: d/dx [f(g(x))] = f'(g(x)) · g'(x).

2.Can we use the derivative of f(g(x)) in real life?

Yes, derivatives of composite functions are used in various fields such as physics, engineering, and economics to model and analyze complex systems and phenomena.

3.Is it possible to take the derivative of f(g(x)) if g(x) is not differentiable?

No, if g(x) is not differentiable at certain points, the derivative of f(g(x)) is undefined at those points.

4.What rule is used to differentiate f(g(x))?

We use the chain rule to differentiate composite functions like f(g(x)): d/dx [f(g(x))] = f'(g(x)) · g'(x).

5.Are the derivatives of f(g(x)) and f(x) · g(x) the same?

No, they are different. The derivative of f(g(x)) is given by the chain rule, while the derivative of f(x) · g(x) is found using the product rule.

6.Can we find the derivative of f(g(x)) if f(x) is a constant?

If f(x) is a constant function, its derivative is zero, making the derivative of f(g(x)) zero regardless of g(x).

Important Glossaries for the Derivative of f(g(x))

Chain Rule: A rule for differentiating composite functions, expressed as d/dx [f(g(x))] = f'(g(x)) · g'(x). Composite Function: A function formed by applying one function to the results of another, denoted as f(g(x)). Higher-Order Derivative: Derivatives obtained by differentiating a function multiple times, such as the second, third, or nth derivative. Differentiable: A function is differentiable at a point if it has a defined derivative at that point. Rate of Change: The rate at which a function changes concerning its variable, often determined by its derivative.

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Jaskaran Singh Saluja

About the Author

Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.

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