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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 3 - DerivativesProblem 3.4.12a
Chapter 3, Problem 3.4.12a

7–14. Find the derivative the following ways:
a. Using the Product Rule (Exercises 7–10) or the Quotient Rule (Exercises 11–14). Simplify your result.
g(s) = 4s³ - 8s² +4s / 4s

Verified step by step guidance
1
Step 1: Identify the function g(s) = \(\frac{4s^3 - 8s^2 + 4s}{4s}\). This is a rational function, so we will use the Quotient Rule to find its derivative.
Step 2: Recall the Quotient Rule, which states that if you have a function h(s) = \(\frac{u(s)}{v(s)}\), then the derivative h'(s) is given by \(\frac{u'(s)v(s) - u(s)v'(s)}{(v(s))^2}\).
Step 3: Assign u(s) = 4s^3 - 8s^2 + 4s and v(s) = 4s. Calculate the derivatives: u'(s) = \(\frac{d}{ds}\)(4s^3 - 8s^2 + 4s) and v'(s) = \(\frac{d}{ds}\)(4s).
Step 4: Compute u'(s) = 12s^2 - 16s + 4 and v'(s) = 4. Substitute these into the Quotient Rule formula: g'(s) = \(\frac{(12s^2 - 16s + 4)(4s) - (4s^3 - 8s^2 + 4s)(4)}{(4s)^2}\).
Step 5: Simplify the expression for g'(s) by expanding the terms in the numerator and then combining like terms. Finally, simplify the entire expression by dividing each term by (4s)^2.

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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 u(s) and v(s) are two differentiable functions, the derivative of their product is given by u'v + uv'. This rule is essential when dealing with functions that are multiplied together, allowing for the correct application of differentiation.
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Quotient Rule

The Quotient Rule is used to differentiate a function that is the quotient of two other functions. If u(s) and v(s) are differentiable functions, the derivative of their quotient is given by (u'v - uv') / v². This rule is particularly important when the function is expressed as a fraction, ensuring accurate differentiation while considering the relationship between the numerator and denominator.
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The Quotient Rule

Simplification of Derivatives

After applying the Product or Quotient Rule, it is often necessary to simplify the resulting expression. This involves combining like terms, factoring, or reducing fractions to make the derivative easier to interpret and use. Simplification is a crucial step in calculus, as it helps clarify the behavior of the function and its rate of change.
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Related Practice
Textbook Question

Find an equation of the line tangent to the following curves at the given value of x.

y = 4 sin x cos x; x = π/3

Textbook Question

21–30. Derivatives

a. Use limits to find the derivative function f' for the following functions f.

f(x) = 1/x+1; a = -1/2;5

Textbook Question

Use the definition of the derivative to determine d/dx(ax²+bx+c), where a, b, and c are constants.

Textbook Question

Derivatives using tables Let h(x)=f(g(x))h(x)=f(g(x)) and p(x)=g(f(x))p(x)=g(f(x)). Use the table to compute the following derivatives.

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a. h′(3)h^{\(\prime\)}\(\left\)(3\(\right\))

Textbook Question

60–62. {Use of Tech} Multiple tangent lines Complete the following steps. <IMAGE>

a. Find equations of all lines tangent to the curve at the given value of x.

x+y³−y=1; x=1

Textbook Question

45–50. Tangent lines Carry out the following steps. <IMAGE>

a. Verify that the given point lies on the curve.

x⁴-x²y+y⁴=1; (−1, 1)