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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 3, Problema 3.23

Find the derivatives of the functions in Exercises 1–42.


𝔂 = x⁻¹/² sec (2x)²

Guida verificata passo dopo passo
1
Identify the function components: The given function is 𝔂 = x-1/2 sec((2x)2). This is a product of two functions: u(x) = x-1/2 and v(x) = sec((2x)2).
Apply the product rule: The derivative of a product of two functions u(x) and v(x) is given by (uv)' = u'v + uv'.
Differentiate u(x): The derivative of u(x) = x-1/2 is u'(x) = -1/2 * x-3/2.
Differentiate v(x): To find v'(x), use the chain rule. The derivative of sec(z) is sec(z)tan(z), where z = (2x)2. First, find the derivative of z with respect to x, which is dz/dx = 4x. Then, v'(x) = sec((2x)2)tan((2x)2) * 4x.
Combine the results: Substitute u(x), u'(x), v(x), and v'(x) into the product rule formula to find the derivative of 𝔂. Simplify the expression to obtain the final derivative.

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Derivative Rules

Understanding the rules of differentiation, such as the product rule, quotient rule, and chain rule, is essential for finding derivatives of complex functions. The product rule is used when differentiating products of functions, while the chain rule is necessary for composite functions. Mastery of these rules allows for systematic and accurate differentiation.
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Trigonometric Functions

The function sec(2x) is a trigonometric function, specifically the secant function, which is the reciprocal of the cosine function. Knowing the derivatives of trigonometric functions, such as sec(x), is crucial for differentiating expressions involving them. The derivative of sec(x) is sec(x)tan(x), which will be applied in this context.
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The power rule is a fundamental concept in calculus that states if f(x) = x^n, then f'(x) = n*x^(n-1). This rule is particularly useful for differentiating functions with exponents, such as x^(-1/2) in the given function. Applying the power rule correctly is vital for simplifying the differentiation process.
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