Derivatives of products and quotients Find the derivative of the following functions by first expanding or simplifying the expression. Simplify your answers.
y = 12s³-8s²+12s/4s

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Step 1: Begin by simplifying the expression \( y = \frac{12s^3 - 8s^2 + 12s}{4s} \). This involves dividing each term in the numerator by the denominator \( 4s \).

Step 2: Simplify each term separately: \( \frac{12s^3}{4s} = 3s^2 \), \( \frac{-8s^2}{4s} = -2s \), and \( \frac{12s}{4s} = 3 \).

Step 3: Rewrite the simplified expression as \( y = 3s^2 - 2s + 3 \).

Step 4: Differentiate the simplified expression term by term. Use the power rule \( \frac{d}{ds}[s^n] = ns^{n-1} \) for each term.

Step 5: Apply the power rule: The derivative of \( 3s^2 \) is \( 6s \), the derivative of \( -2s \) is \( -2 \), and the derivative of \( 3 \) is \( 0 \). Combine these to find the derivative of the function.

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주요 개념

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Derivatives

A derivative represents the rate of change of a function with respect to a variable. It is a fundamental concept in calculus that allows us to determine how a function behaves at any given point. The derivative can be computed using various rules, including the power rule, product rule, and quotient rule, depending on the form of the function.

Product and Quotient Rules

The product rule and quotient rule are techniques used to differentiate functions that are products or quotients of two or more 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 quotient rule, on the other hand, provides a method for differentiating a quotient of two functions, ensuring that the denominator is squared in the final expression.

Simplification of Expressions

Simplifying expressions is a crucial step in calculus that often makes differentiation easier. This involves combining like terms, factoring, or reducing fractions to their simplest form. In the context of the given function, simplifying the expression before taking the derivative can lead to a more straightforward calculation and a clearer final result.

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