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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 10a

Find the derivative the following ways:
Using the Product Rule or the Quotient Rule. Simplify your result.
h(z) = (z3 + 4z2 + z)(z - 1)

Verified step by step guidance
1
Step 1: Identify the functions to apply the Product Rule. Here, we have two functions: \( u(z) = z^3 + 4z^2 + z \) and \( v(z) = z - 1 \).
Step 2: Recall the Product Rule formula: \( (uv)' = u'v + uv' \). We need to find the derivatives \( u'(z) \) and \( v'(z) \).
Step 3: Differentiate \( u(z) = z^3 + 4z^2 + z \). Using the power rule, \( u'(z) = 3z^2 + 8z + 1 \).
Step 4: Differentiate \( v(z) = z - 1 \). The derivative is \( v'(z) = 1 \).
Step 5: Substitute \( u(z) \), \( u'(z) \), \( v(z) \), and \( v'(z) \) into the Product Rule formula: \( h'(z) = (3z^2 + 8z + 1)(z - 1) + (z^3 + 4z^2 + z)(1) \). Simplify the expression to find the derivative.

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Key Concepts

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

Product Rule

The Product Rule is a formula used to find the derivative of the product of two functions. If you have two functions, u(z) and v(z), the derivative of their product is given by u'v + uv'. This rule is essential when differentiating expressions where two functions are multiplied together, as it allows for a systematic approach to finding the derivative.
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The Product Rule

Quotient Rule

The Quotient Rule is used to differentiate a function that is the quotient of two other functions. If f(z) = u(z)/v(z), the derivative is given by (u'v - uv')/v^2. This rule is particularly useful when dealing with fractions of functions, ensuring that the derivative accounts for both the numerator and the denominator.
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The Quotient Rule

Simplification of Derivatives

After applying the Product or Quotient Rule, the resulting derivative often needs to be simplified. This involves combining like terms, factoring, or reducing fractions to make the expression more manageable. Simplification is crucial for clarity and for further analysis, such as finding critical points or analyzing the behavior of the function.
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Derivatives