Chapter 5: Problem 24
Show that (a) \(\frac{d}{d x}\left[\int_{x}^{a} f(t) d t\right]=-f(x)\) (b) \(\frac{d}{d x}\left[\int_{g(x)}^{a} f(t) d t\right]=-f(g(x)) g^{\prime}(x) .\)
Short Answer
Expert verified
(a) \(-f(x)\); (b) \(-f(g(x))g'(x)\).
Step by step solution
01
Understanding the Problem for Part (a)
We need to find the derivative with respect to \( x \) of the integral \( \int_{x}^{a} f(t) dt \). This requires using the Fundamental Theorem of Calculus.
02
Applying the Fundamental Theorem of Calculus to Part (a)
The Fundamental Theorem of Calculus states that if \( F \) is an antiderivative of \( f \), then \( \frac{d}{dx} \left[ \int_{x}^{a} f(t) dt \right] = -f(x) \). The negative sign arises because the lower bound of integration, \( x \), varies.
03
Conclusion for Part (a)
Thus, by the Fundamental Theorem of Calculus, \( \frac{d}{dx} \left[ \int_{x}^{a} f(t) dt \right] = -f(x) \) is proven.
04
Understanding the Problem for Part (b)
For part (b), we consider the integral \( \int_{g(x)}^{a} f(t) dt \). We need to differentiate this with respect to \( x \), which involves using both the Fundamental Theorem of Calculus and the chain rule.
05
Applying the Chain Rule for Part (b)
Firstly, consider the integral \( I(x) = \int_{g(x)}^{a} f(t) dt \). The derivative \( \frac{dI}{dx} \) is the composition of functions where the lower limit is a function of \( x \). Using the Fundamental Theorem of Calculus and the chain rule, we have \( \frac{d}{dx}[I(x)] = -f(g(x)) \cdot g'(x) \).
06
Conclusion for Part (b)
Thus, incorporating both the fundamental theorem and the chain rule, the derivative \( \frac{d}{dx} \left[ \int_{g(x)}^{a} f(t) dt \right] = -f(g(x)) g'(x) \) is shown.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivatives
Derivatives are a core concept in calculus, representing the rate of change of a function with respect to its variable. Imagine you are driving a car, and you want to know how fast you’re going at any given moment. The speedometer tells you that rate – that’s your derivative! In mathematical terms, if you have a function \( f(x) \), its derivative, denoted as \( f'(x) \) or \( \frac{df}{dx} \), tells you how \( f \) changes as \( x \) changes.
- Derivative of a Constant: \( \frac{d}{dx}(c) = 0 \) since constants do not change.
- Power Rule: For \( f(x) = x^n \), \( \frac{d}{dx}(x^n) = nx^{n-1} \).
- Product and Quotient Rules: Useful for differentiating products or quotients of functions.
Chain rule
The chain rule is a fundamental tool for finding the derivative of composite functions. It’s like peeling an onion where one layer depends on another. For a function composed of functions, like \( f(g(x)) \), the chain rule helps us differentiate it.
Here's the basic idea: if you have a function \( y = f(u) \) and \( u = g(x) \), then the derivative \( \frac{dy}{dx} \) is the product of the inner and outer functions' derivatives:\[\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\]
Here's the basic idea: if you have a function \( y = f(u) \) and \( u = g(x) \), then the derivative \( \frac{dy}{dx} \) is the product of the inner and outer functions' derivatives:\[\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\]
- Helps to differentiate when the expression is complicated.
- Essential in dealing with functions within other functions.
Antiderivatives
Antiderivatives, often called indefinite integrals, are the 'backwards' of derivatives. If differentiation tells us how to go from a function to its rate of change, antiderivates tell us how to go from a rate of change back to a function. Specifically, if \( F'(x) = f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \).
- Notation: The antiderivative of \( f(x) \) is \( \int f(x) \, dx \).
- It gives a family of functions differing by a constant \( C \), written as \( F(x) + C \).
- Fundamental in finding areas and solving differential equations.