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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
3장, 문제 120a

Let f(x) = x².
a. Show that f(x)−f(y) / x−y = f′(x+y²), for all x≠y.

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1
Start by understanding the problem: We need to show that the difference quotient (f(x) - f(y)) / (x - y) is equal to the derivative of f evaluated at (x + y²).
First, calculate f(x) and f(y) using the given function f(x) = x². This gives us f(x) = x² and f(y) = y².
Substitute these into the difference quotient: (f(x) - f(y)) / (x - y) = (x² - y²) / (x - y).
Recognize that x² - y² is a difference of squares, which can be factored as (x - y)(x + y). Substitute this into the difference quotient to simplify: ((x - y)(x + y)) / (x - y).
Cancel the (x - y) terms in the numerator and denominator, assuming x ≠ y, to get x + y. Now, find the derivative f'(x) = 2x, and evaluate it at x + y²: f'(x + y²) = 2(x + y²). Compare this with the simplified expression x + y to verify the equality.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
2m

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Difference Quotient

The difference quotient is a fundamental concept in calculus that represents the average rate of change of a function over an interval. It is defined as (f(x) - f(y)) / (x - y) for x ≠ y. This expression is crucial for understanding the derivative, as it approaches the instantaneous rate of change as y approaches x.
추천 영상:
06:43
The Quotient Rule

Derivative

The derivative of a function at a point measures how the function's output changes as its input changes. It is denoted as f'(x) and can be interpreted as the slope of the tangent line to the function's graph at that point. In this question, we need to show that the difference quotient equals the derivative of f at a specific point, which involves applying the definition of the derivative.
추천 영상:

Function Composition

Function composition involves combining two functions to create a new function, where the output of one function becomes the input of another. In this context, we need to evaluate f' at (x + y²), which requires understanding how to apply the derivative to a function that is itself a composition of variables. This concept is essential for manipulating and simplifying expressions involving derivatives.
추천 영상:
3:48
Evaluate Composite Functions - Special Cases