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Chapter 11: Problem 16

Obtain the derivative \(d y / d x\) and state the rules that you use. HINT [See Example 2.] $$ y=4 x^{-1}-2 x-10 $$

Short Answer

Expert verified
The derivative of the function \(y = 4x^{-1} - 2x - 10\) with respect to \(x\) is: \(y' = \frac{-4}{x^2} + 2\). The rules used are the power rule and the constant rule in differentiation.

Step by step solution

01

Rewrite the function

First, rewrite the function with a negative exponent as a fraction: \( y = 4x^{-1} - 2x - 10 = \frac{4}{x} - 2x -10 \)
02

Apply the power rule for differentiation

Apply the power rule for differentiation for each term. The power rule states that for a function \(f(x) = x^k\), its derivative is \(f'(x) = kx^{k-1}\). \( y' = \frac{d}{dx}(\frac{4}{x}) - \frac{d}{dx}(2x) -\frac{d}{dx}(10) \)
03

Derivative of the first term

Find the derivative of the first term: \( \frac{d}{dx}(\frac{4}{x}) = 4\frac{d}{dx}(x^{-1}) = 4(-1)x^{-1-1} = -4x^{-2}\)
04

Derivative of the second term

Find the derivative of the second term: \( \frac{d}{dx}(2x) = 2\frac{d}{dx}(x) = 2(x^{1})' = 2(1)x^{1-1} = 2x^{0} = 2 \)
05

Derivative of the third term

Find the derivative of the third term, which is constant. Using constant rule, the derivative of a constant is zero: \( \frac{d}{dx}(10) = 0 \)
06

Combine the derivatives

Combine the derivatives of all the terms in the function to get the final result: \( y' = -4x^{-2} + 2 - 0 = -4x^{-2} + 2 \)
07

Rewrite the derivative

Now rewrite the derivative back in terms of a fraction: \( y' = \frac{-4}{x^2} + 2 \) So, the derivative of the function \(y = 4x^{-1} - 2x - 10\) with respect to \(x\) is: \( \boxed{y' = \frac{-4}{x^2} + 2} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule for Differentiation
The power rule is a fundamental principle in calculus used to find the derivative of a function that contains a term raised to a power. This rule states that if you have a function of the form f(x) = x^k, where k is a real number, then its derivative is f'(x) = kx^{k-1}. Put simply, you bring down the exponent as a coefficient in front of the variable, and then subtract one from the exponent.

To apply this to a term like 2x in the given exercise, we identify k = 1 since 2x is the same as 2x^1. Applying the power rule, the derivative is 2(1)x^{1-1} which simplifies to 2 because any nonzero number to the power of zero equals one. In terms of negative exponents, for a term like x^{-1}, the power rule gives us a derivative of (-1)x^{-1-1} or -x^{-2} after simplifying. The power rule is a swift and efficient method to differentiate polynomials and any function that can be expressed as a power of x.
Constant Rule in Calculus
In calculus, the constant rule is an essential concept that states the derivative of a constant is always zero. This is because a constant does not change, thus, it has no rate of change, which is what a derivative measures.

Applying this to the given exercise, we have a constant term, -10. According to the constant rule, the derivative of this term with respect to x is zero: d/dx(10) = 0. This rule simplifies the process of differentiation because we can simply ignore constant terms when differentiating a function, as they will not contribute to the derivative.
Negative Exponents
A negative exponent indicates that the number is a reciprocal. For instance, x^{-1} is the same as 1/x. When differentiating terms with negative exponents, we apply the power rule as usual, but we must be cautious with the signs.

In the exercise, the term 4x^{-1} contains a negative exponent. To differentiate this term, we first transform the term with the negative exponent into a fraction and then apply the power rule. The derivative of 4x^{-1} becomes 4(-1)x^{-1-1}, which simplifies to -4x^{-2}. Then, we can express this negative exponent as a positive one in the denominator, giving us -4/x^2 as part of the final derivative. Understanding how to manage negative exponents is crucial in calculus, especially when dealing with rational functions and variables in the denominator.

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