Ch. 3 - Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo

Hass 15th Edition

Ch. 3 - Derivatives

Problema 3.110

Capitolo 3, Problema 3.110

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Find the linearization of ƒ(x) = 2/ (1 - x) + √1 + x - 3.1 at x = 0.

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Identify the function to be linearized: \( f(x) = \frac{2}{1-x} + \sqrt{1+x} - 3.1 \).

Recall the formula for the linearization of a function at a point \( a \): \( L(x) = f(a) + f'(a)(x-a) \). Here, \( a = 0 \).

Calculate \( f(0) \) by substituting \( x = 0 \) into the function: \( f(0) = \frac{2}{1-0} + \sqrt{1+0} - 3.1 \).

Find the derivative \( f'(x) \). Use the sum rule and differentiate each term separately: \( f'(x) = \frac{d}{dx}\left(\frac{2}{1-x}\right) + \frac{d}{dx}(\sqrt{1+x}) \).

Evaluate \( f'(0) \) by substituting \( x = 0 \) into the derivative. Then, use the linearization formula to find \( L(x) = f(0) + f'(0)(x-0) \).

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Linearization

Linearization is the process of approximating a function near a specific point using its tangent line. This involves finding the function's value and its derivative at that point. The linearization formula is given by L(x) = f(a) + f'(a)(x - a), where 'a' is the point of tangency. This technique is useful for simplifying complex functions for easier analysis.

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Linearization

Derivative

The derivative of a function measures the rate at which the function's value changes as its input changes. It is defined as the limit of the average rate of change of the function over an interval as the interval approaches zero. In the context of linearization, the derivative at a point provides the slope of the tangent line, which is essential for constructing the linear approximation.

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Derivatives

Function Evaluation

Function evaluation involves calculating the output of a function for a given input. In this case, we need to evaluate the function ƒ(x) at x = 0 to find the point of tangency for linearization. This step is crucial as it provides the y-coordinate of the tangent line, which, along with the slope from the derivative, defines the linear approximation of the function.

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Evaluating Composed Functions

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The diameter of a sphere is measured as 100 ± 1 cm and the volume is calculated from this measurement. Estimate the percentage error in the volume calculation.

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Second Derivatives

In Exercises 19–26, use implicit differentiation to find dy/dx and then d²y/dx². Write the solutions in terms of x and y only.

x²/³ + y²/³ = 1

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In Exercises 53 and 54, find dr/ds.

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