Ch. 3 - Derivatives

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 3 - Derivatives

Problem 3.14

Chapter 3, Problem 3.14

9–61. Evaluate and simplify y'.

y = (2x−3)x^3/2

Verified step by step guidance

Step 1: Identify the function y = (2x - 3)x^{3/2}. This is a product of two functions, so we will use the product rule to differentiate it.

Step 2: Recall the product rule for differentiation: if y = u(x)v(x), then y' = u'(x)v(x) + u(x)v'(x). Here, let u(x) = 2x - 3 and v(x) = x^{3/2}.

Step 3: Differentiate u(x) = 2x - 3. The derivative u'(x) is 2, since the derivative of 2x is 2 and the derivative of a constant is 0.

Step 4: Differentiate v(x) = x^{3/2}. Use the power rule for differentiation: if v(x) = x^n, then v'(x) = nx^{n-1}. Here, n = 3/2, so v'(x) = (3/2)x^{1/2}.

Step 5: Apply the product rule: y' = u'(x)v(x) + u(x)v'(x). Substitute u'(x), v(x), u(x), and v'(x) into this formula to find y'.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. The derivative represents the rate of change of the function with respect to its variable. In this case, we need to apply differentiation rules to the given function y = (2x−3)x^(3/2) to find y'.

Product Rule

The Product Rule is a specific rule used in differentiation when dealing with the product of two functions. It states that if you have two functions u(x) and v(x), the derivative of their product is given by u'v + uv'. In the context of the given function, we will apply the Product Rule to differentiate the two components: (2x−3) and x^(3/2).

Simplification

Simplification in calculus involves reducing an expression to its simplest form after differentiation. This may include combining like terms, factoring, or reducing fractions. After finding the derivative y', it is essential to simplify the expression to make it easier to interpret and use in further calculations.

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