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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 25

Evaluate the derivative of the following functions.
f(x) = x2 + 2x3 cot-1 x - ln (1 + x2)

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To find the derivative of the function f(x) = x^2 + 2x^3 * cot^(-1)(x) - ln(1 + x^2), we will apply the rules of differentiation to each term separately.
Start with the first term, x^2. The derivative of x^2 with respect to x is 2x, using the power rule.
Next, consider the term 2x^3 * cot^(-1)(x). This is a product of two functions, so we will use the product rule: if u(x) = 2x^3 and v(x) = cot^(-1)(x), then the derivative is u'(x)v(x) + u(x)v'(x). Calculate u'(x) = 6x^2 and v'(x) = -1/(1 + x^2).
For the term -ln(1 + x^2), use the chain rule. The derivative of ln(u) is 1/u * du/dx. Here, u = 1 + x^2, so du/dx = 2x. Therefore, the derivative is -1/(1 + x^2) * 2x.
Combine the derivatives of each term to find the derivative of the entire function: f'(x) = 2x + (6x^2 * cot^(-1)(x) + 2x^3 * (-1/(1 + x^2))) - (2x/(1 + x^2)).

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주요 개념

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Derivative

The derivative of a function measures how the function's output value changes as its input value changes. It is defined as the limit of the average rate of change of the function over an interval as the interval approaches zero. Derivatives are fundamental in calculus for understanding rates of change and are denoted as f'(x) or df/dx.
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Product and Chain Rules

The product rule and chain rule are essential techniques for finding derivatives of more complex functions. The product rule states that the derivative of a product of two functions is the first function times the derivative of the second plus the second function times the derivative of the first. The chain rule is used when differentiating composite functions, stating that the derivative of f(g(x)) is f'(g(x)) * g'(x).
추천 영상:
05:18
The Product Rule

Logarithmic Differentiation

Logarithmic differentiation is a technique used to differentiate functions that are products or quotients of variables raised to powers. By taking the natural logarithm of both sides of the equation, it simplifies the differentiation process, especially when dealing with products or powers. This method is particularly useful for functions involving logarithms, as seen in the given function with ln(1 + x²).
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Logarithmic Differentiation
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