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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 3, Problema 29

Find and simplify the derivative of the following functions.
y = (3t−1)(2t−2)-1

Guida verificata passo dopo passo
1
Step 1: Identify the function y = (3t - 1)(2t - 2)^{-1} as a product of two functions, u(t) = 3t - 1 and v(t) = (2t - 2)^{-1}.
Step 2: Use the product rule for differentiation, which states that if y = u(t)v(t), then y' = u'(t)v(t) + u(t)v'(t).
Step 3: Differentiate u(t) = 3t - 1 to find u'(t). The derivative of 3t is 3, and the derivative of -1 is 0, so u'(t) = 3.
Step 4: Differentiate v(t) = (2t - 2)^{-1} using the chain rule. Rewrite v(t) as (2t - 2)^{-1} = (2t - 2)^{-1} = (2t - 2)^{-1}. The derivative of (2t - 2)^{-1} is -1(2t - 2)^{-2} times the derivative of (2t - 2), which is 2.
Step 5: Substitute u(t), u'(t), v(t), and v'(t) into the product rule formula to find y'. Simplify the expression to obtain the derivative of y.

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Derivative

The derivative of a function measures how the function's output value changes as its input changes. It is a fundamental concept in calculus, representing the slope of the tangent line to the curve at any given point. The derivative can be computed using various rules, such as the power rule, product rule, and quotient rule, depending on the form of the function.
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Product Rule

The product rule is a formula used to find the derivative of the product of two functions. It states that if you have two functions u(t) and v(t), the derivative of their product is given by u'v + uv'. This rule is essential when differentiating functions that are multiplied together, as in the given function y = (3t−1)(2t−2)⁻¹.
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The Product Rule

Quotient Rule

The quotient rule is used to differentiate functions that are expressed as the ratio of two other functions. If y = u(t)/v(t), the derivative is given by (u'v - uv')/v². This rule is particularly relevant for the function in the question, as it involves a term raised to a negative exponent, which can be interpreted as a quotient.
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The Quotient Rule