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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 3 - DerivativesProblem 3.3.63b
Chapter 3, Problem 3.3.63b

Generalizing the Product Rule The Derivative Product Rule gives the formula


d/dx (uv) = u (dv/dx) + (du/dx) v


for the derivative of the product uv of two differentiable functions of x.


b. What is the formula for the derivative of the product u₁u₂u₃u₄ of four differentiable functions of x?

Verified step by step guidance
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To find the derivative of the product of four functions, u₁u₂u₃u₄, we can extend the product rule. The product rule for two functions states that the derivative of a product uv is given by d/dx (uv) = u (dv/dx) + (du/dx) v.
For four functions, we apply the product rule iteratively. First, consider the product of the first two functions, u₁u₂, and the product of the last two functions, u₃u₄. We can treat these as two separate products and apply the product rule to each.
The derivative of u₁u₂u₃u₄ can be expressed as: d/dx (u₁u₂u₃u₄) = (d/dx (u₁u₂)) u₃u₄ + u₁u₂ (d/dx (u₃u₄)).
Now, apply the product rule to each part: d/dx (u₁u₂) = u₁ (du₂/dx) + (du₁/dx) u₂ and d/dx (u₃u₄) = u₃ (du₄/dx) + (du₃/dx) u₄.
Substitute these derivatives back into the expression: d/dx (u₁u₂u₃u₄) = [u₁ (du₂/dx) + (du₁/dx) u₂] u₃u₄ + u₁u₂ [u₃ (du₄/dx) + (du₃/dx) u₄].

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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 fundamental theorem in calculus that provides a method for differentiating the product of two functions. It states that the derivative of the product of two functions u and v is given by d/dx(uv) = u(dv/dx) + v(du/dx). This rule can be extended to more than two functions, which is essential for solving problems involving multiple products.
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Higher-Order Product Rule

The Higher-Order Product Rule generalizes the Product Rule to the differentiation of products involving more than two functions. For three functions, the derivative can be expressed as d/dx(u₁u₂u₃) = u₁(u₂' u₃ + u₂ u₃') + u₂(u₁' u₃ + u₁ u₃') + u₃(u₁' u₂ + u₁ u₂'). This pattern continues, allowing for systematic differentiation of products of any number of functions.
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Differentiability

Differentiability is a key concept in calculus that indicates whether a function has a derivative at a given point. A function is differentiable at a point if it is continuous there and its derivative exists. This property is crucial when applying the Product Rule, as it ensures that the functions involved can be differentiated, allowing for the application of the rule to find the derivative of their product.
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Related Practice
Textbook Question

Consider the function


f(x) = { x² cos(2/x), x ≠ 0

0, x = 0


b. Determine f' for x ≠ 0.

1
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Textbook Question

Temperatures in Fairbanks, Alaska The graph in the accompanying figure shows the average Fahrenheit temperature in Fairbanks, Alaska, during a typical 365-day year. The equation that approximates the temperature on day x is


y = 37 sin[(2π/365)(x − 101)] + 25


and is graphed in the accompanying figure.


b. About how many degrees per day is the temperature increasing when it is increasing at its fastest?


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Textbook Question

Right circular cylinder The total surface area S of a right circular cylinder is related to the base radius r and height h by the equation S = 2πr² + 2πrh.


b. How is dS/dt related to dh/dt if r is constant?

Textbook Question

Computer Explorations


Use a CAS to perform the following steps in Exercises 55–62.


b. Using implicit differentiation, find a formula for the derivative dy/dx and evaluate it at the given point P.


x√(1 + 2y) + y = x², P(1,0)

Textbook Question

Area The area A of a triangle with sides of lengths a and b enclosing an angle of measure θ is

A = (1/2) ab sinθ.


b. How is dA/dt related to dθ/dt and da/dt if only b is constant?

Textbook Question

b. Show that


f(x) = { x² sin(1/x), x ≠ 0

0, x = 0

is differentiable at x = 0 and find f′(0).