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

Short Answer

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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 by step solution

01

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
02

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.
03

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
04

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
05

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).

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