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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Briggs 3rd EditionCh. 3 - Derivatives문제 3.4.51
3장, 문제 3.4.51

Derivatives Find and simplify the derivative of the following functions.
h(w) = w⁵/³ / w⁵/³+1

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Step 1: Identify the function to differentiate. The function given is \( h(w) = \frac{w^{5/3}}{w^{5/3} + 1} \).
Step 2: Recognize that this is a quotient of two functions, so apply the Quotient Rule. The Quotient Rule states that if you have a function \( \frac{u}{v} \), its derivative is \( \frac{u'v - uv'}{v^2} \). Here, \( u = w^{5/3} \) and \( v = w^{5/3} + 1 \).
Step 3: Differentiate the numerator \( u = w^{5/3} \). Using the power rule, \( u' = \frac{5}{3}w^{2/3} \).
Step 4: Differentiate the denominator \( v = w^{5/3} + 1 \). The derivative of \( v \) is \( v' = \frac{5}{3}w^{2/3} \) since the derivative of a constant is zero.
Step 5: Substitute \( u' \), \( v \), \( u \), and \( v' \) into the Quotient Rule formula: \( h'(w) = \frac{\left(\frac{5}{3}w^{2/3}\right)(w^{5/3} + 1) - (w^{5/3})\left(\frac{5}{3}w^{2/3}\right)}{(w^{5/3} + 1)^2} \). Simplify the expression to find the derivative.

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주요 개념

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Derivatives

A derivative represents the rate at which a function changes at any given point. It is a fundamental concept in calculus that measures how a function's output value changes as its input value changes. The derivative can be interpreted as the slope of the tangent line to the curve of the function at a specific point.
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Derivatives

Quotient Rule

The Quotient Rule is a formula used to find the derivative of a function that is the ratio of two other functions. If you have a function in the form f(x) = g(x)/h(x), the derivative is given by f'(x) = (g'(x)h(x) - g(x)h'(x)) / (h(x))². This rule is essential for differentiating functions like the one in the question, where one function is divided by another.
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The Quotient Rule

Simplification of Derivatives

After finding the derivative of a function, simplification is often necessary to express the result in a more manageable form. This may involve factoring, reducing fractions, or combining like terms. Simplifying the derivative helps in understanding the behavior of the function and makes it easier to analyze critical points and concavity.
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Derivatives
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