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Find \(F^{\prime}(x)\). Assume that all functions are differentiable. $$ F(x)=f\left(x^{2}+1\right)+g\left(x^{2}-1\right) $$

Short Answer

Expert verified
The short answer is: \(F'(x) = 2x(f'(x^2+1) + g'(x^2-1))\)

Step by step solution

01

Apply the chain rule

The chain rule states that (f(g(x)))' = f'(g(x)) * g'(x). We need to apply the chain rule to both functions f(x²+1) and g(x²-1). \(F'(x) = (f(x^2+1))' + (g(x^2-1))'\)
02

Find the derivative of f(x²+1)

According to the chain rule, we need to find the derivatives of the outer function f and inner function x²+1. Outer function: f(x) Inner function: x²+1 Derivative of outer function f'(x) with respect to x. Derivative of inner function (x²+1)' = 2x \(F'(x) = (f'(x^2+1) * 2x) + (g(x^2-1))'\)
03

Find the derivative of g(x²-1)

Similarly, use the chain rule to find the derivatives of the outer and inner functions. Outer function: g(x) Inner function: x²-1 Derivative of outer function g'(x) with respect to x. Derivative of inner function (x²-1)' = 2x \(F'(x) = (f'(x^2+1) * 2x) + (g'(x^2-1) * 2x)\)
04

Simplify the expression

Now, we can simplify the expression and write the final result. \(F'(x) = 2x(f'(x^2+1) + g'(x^2-1))\) Thus, the derivative of F(x) is: \(F'(x) = 2x(f'(x^2+1) + g'(x^2-1))\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
A derivative represents the rate at which a function is changing at any given point. It is an essential concept in calculus, as it tells us how a function behaves as its input changes. Imagine you're riding a bike up a hill. The steepness or the angle of the hill at any point is like the derivative; it shows how quickly the elevation changes as you move forward. In mathematical terms, the derivative of a function at a point provides the slope of the tangent line to the curve of the function at that point. It is often written as \( f'(x) \) or \( \frac{df}{dx} \). Derivatives simplify complex problems in physics, engineering, and economics by pinpointing exact rates of change and behavior trends. In the context of our problem, the derivative \( F'(x) \) gives us insight into how the combined changes of the functions \( f \) and \( g \) respond to alterations in \( x \).
Differentiable Functions
A function is considered differentiable at a point if it has a derivative there. It's like being 'smooth' without any gaps or sharp points, similar to a perfectly smooth ski slope. Not every function has a derivative. For a function to be differentiable, it must be continuous at that point. This means there are no breaks, jumps, or holes in the graph of that function. Further along this ski analogy, if there's a sudden cliff, the slope and direction—the derivative—become undefined at that point. In our exercise, functions \( f \) and \( g \) are assumed to be differentiable, meaning they're smooth everywhere on their domain. In calculus, understanding which functions are differentiable allows us to apply tactics such as the chain rule effectively, ensuring accurate computation of rates of change.
Outer and Inner Functions
When working with composite functions, it's crucial to identify the outer and inner functions to apply rules like the chain rule effectively. The outer function is the one that is applied last, and the inner function is the one applied first. For example, in the expression \( f(x^2 + 1) \), \( f \) is the outer function, while \( x^2 + 1 \) is the inner function. Understanding these functions helps break down complex expressions into manageable parts. The chain rule essentially tells us how to handle derivatives of such nested compositions by first differentiating the outer function and then multiplying it by the derivative of the inner function. In our original exercise, recognizing \( x^2 + 1 \) and \( x^2 - 1 \) as inner functions and \( f \) and \( g \) as outer functions is essential for correctly applying the chain rule and solving the derivative challenge.
Calculus
Calculus is the branch of mathematics that studies how things change. It provides the tools to calculate trends, slopes of curves, areas under curves, and much more. The two fundamental tools of calculus are derivatives and integrals. In this context, we've been exploring derivatives. Calculus breaks down complex problems by examining the small and accumulating their effects to understand the whole picture. This is done using limits, which allow us to find the exact slope of a curve at a point or the total area under a curve—things that algebra alone could not handle. By understanding derivatives through calculus concepts, we can predict how changes in one quantity affect another, a powerful tool across scientific disciplines and real-world applications.

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