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Use \(f^{\prime}(x)=\lim _{h \rightarrow 0}[f(x+h)-f(x)] / h\) to find the derivative at \(x\). \(h(x)=\frac{2}{x}\)

Short Answer

Expert verified
The derivative of \( h(x) = \frac{2}{x} \) is \( h'(x) = \frac{-2}{x^2} \).

Step by step solution

01

Write the Difference Quotient

We begin by writing the difference quotient for the function \( h(x) = \frac{2}{x} \). This will help us in finding the derivative: \[ \frac{h(x+h) - h(x)}{h} = \frac{\frac{2}{x+h} - \frac{2}{x}}{h} \]
02

Simplify the Difference

Combine the fractions in the numerator of the difference quotient by getting a common denominator:\[\frac{2}{x+h} - \frac{2}{x} = \frac{2x - 2(x+h)}{x(x+h)} = \frac{2x - 2x - 2h}{x(x+h)} = \frac{-2h}{x(x+h)}\]
03

Insert Simplified Difference into the Quotient and Cancel Terms

Replace the numerator in the difference quotient with the simplified expression from Step 2 and cancel \( h \):\[\frac{\frac{-2h}{x(x+h)}}{h} = \frac{-2h}{hx(x+h)} = \frac{-2}{x(x+h)}\]
04

Take the Limit as \(h\) Approaches Zero

Take the limit of the difference quotient as \( h \) approaches 0:\[\lim_{h \to 0} \frac{-2}{x(x+h)} = \frac{-2}{x^2}\]
05

State the Derivative

After evaluating the limit, we find the derivative of the function \( h(x) = \frac{2}{x} \): \[ h'(x) = \frac{-2}{x^2} \]

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Key Concepts

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

Difference Quotient
The difference quotient is a crucial concept when it comes to finding derivatives in calculus. It helps in determining the rate at which a function changes. The generic formula for a difference quotient of a function \( f(x) \) is:
  • \( \frac{f(x+h) - f(x)}{h} \)
When \( h \) is a tiny increment added to \( x \), the difference \( f(x+h) - f(x) \) essentially measures the change in the function over that small span. Hence, the quotient represents the average rate of change over that interval.

For the function \( h(x) = \frac{2}{x} \), the difference quotient would look like:
  • \( \frac{\frac{2}{x+h} - \frac{2}{x}}{h} \)
This expression needs simplification in order to find the function's derivative effectively.
Rational Functions
Rational functions are expressions formed by the ratio of two polynomials. A basic understanding of their structure helps in manipulating expressions for calculus tasks. In our case, \( h(x) = \frac{2}{x} \) is a simple rational function. The key characteristics of rational functions include:
  • Being defined as \( \frac{P(x)}{Q(x)} \) where both \( P(x) \) and \( Q(x) \) are polynomials.
  • The domain excludes values where \( Q(x) = 0 \) since division by zero is undefined.
  • They often require the application of algebraic methods for simplification, especially when working with derivatives.
To derive \( h(x) = \frac{2}{x} \), combining fractions by finding a common denominator is essential. This process yields effective manipulation of the difference quotient.
Limit Process
The limit process is the heart of calculus, providing a method to assess the behavior of functions as they approach specific points. When computing the derivative, we take the limit of the difference quotient as \( h \) becomes infinitesimally small, specifically:
  • \( \lim_{h \to 0} \left( \frac{-2}{x(x+h)} \right) \)
Limits explore what happens to the function values as \( h \) approaches zero. In this context, when \( h \to 0 \), the term \( x+h \) simplifies indirectly to \( x \), helping refine our expression to:
  • \( \lim_{h \to 0} \frac{-2}{x(x+h)} = \frac{-2}{x^2} \)
This result represents the derivative of the original function, \( h(x) = \frac{2}{x} \), showing how function values change at \( x \).

Thus, the limit process is not just a mathematical tool, but a core underpinning of calculus that gives precise insight into function dynamics.

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