Problem 2 - Calculus Early Transcendentals

Question

How do you find the derivative of the quotient of two functions that are differentiable at a point?

Step-by-Step Solution

Answer

Question: Using the Quotient Rule, find the derivative of the function h(x) = (3x^2 + x) / (x^2 - 1). Answer: h'(x) = (-2x^4 + x^2 - 6x - 1) / (x^4 - 2x^2 + 1)

Step 1: Recall the Quotient Rule formula

The Quotient Rule states that if we have two functions, f(x) and g(x), and we want to find the derivative of their quotient h(x) = f(x) / g(x), then the derivative h'(x) can be found using the formula: h'(x) = (g(x) * f'(x) - f(x) * g'(x)) / (g(x))^2

Step 2: Identify the two functions

Identify the two functions f(x) and g(x) which form the quotient. For instance, if given the problem h(x) = (3x^2 + x) / (x^2 - 1), then f(x) = 3x^2 + x and g(x) = x^2 - 1.

Step 3: Find their derivatives

Use the rules of differentiation to find the derivatives of the functions f(x) and g(x) separately. Continuing with the example given in Step 2: f'(x) = d/dx (3x^2 + x) = 6x + 1 g'(x) = d/dx (x^2 - 1) = 2x

Step 4: Apply the Quotient Rule formula

Substitute the functions f(x), g(x) and their derivatives f'(x), g'(x) into the Quotient Rule formula to find h'(x): h'(x) = ((x^2 - 1) * (6x + 1) - (3x^2 + x) * (2x)) / (x^2 - 1)^2

Step 5: Simplify the expression

Simplify the resulting expression to find the derivative in its simplest form. In the given example: h'(x) = (6x^3 + x^2 - 6x - 1 - 6x^3 - 2x^4) / (x^4 - 2x^2 + 1) = (-2x^4 + x^2 - 6x - 1) / (x^4 - 2x^2 + 1) Now we have found the derivative of the quotient of the two functions, h'(x) = (-2x^4 + x^2 - 6x - 1) / (x^4 - 2x^2 + 1).