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Chapter 2: Problem 7

In each case, find the derivative \(\frac{d y}{d x}\) from first principles. a. \(y=6-7 x\) b. \(y=\frac{x+1}{x-1}\) c. \(y=3 x^{2}\)

Short Answer

Expert verified
a. \(\frac{d y}{d x} = -7\); b. \(\frac{d y}{d x} = \frac{-2}{(x-1)^2}\); c. \(\frac{d y}{d x} = 6x\).

Step by step solution

01

Introduction to Differentiation from First Principles

Differentiating a function from first principles involves finding the limit of the difference quotient as the change in the input approaches zero. The difference quotient for a function \(y=f(x)\) is given by \( \frac{f(x+h) - f(x)}{h} \). The derivative is \( \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \).
02

Derivative of a Linear Function

For the function \(y = 6 - 7x\), find \(f(x+h)\):\(f(x+h) = 6 - 7(x+h) = 6 - 7x - 7h\).Compute the difference quotient:\(\frac{f(x+h) - f(x)}{h} = \frac{(6 - 7x - 7h) - (6 - 7x)}{h} = \frac{-7h}{h} = -7\).Take the limit as \(h\to 0\):\(\lim_{h\to 0} (-7) = -7\).Therefore, \(\frac{d y}{d x} = -7\).
03

Derivative of a Rational Function

For \(y=\frac{x+1}{x-1}\), find \(f(x+h)\):\(f(x+h) = \frac{(x+h)+1}{(x+h)-1}\).Difference quotient:\(\frac{f(x+h) - f(x)}{h} = \frac{\frac{x+h+1}{x+h-1} - \frac{x+1}{x-1}}{h}\).Combine the fractions:\( \frac{(x+h+1)(x-1) - (x+1)(x+h-1)}{h(x+h-1)(x-1)} \).Simplify the numerator:Expand and simplify: \((x+h+1)(x-1) - (x+1)(x+h-1) = x^2 + xh + x - xh - h - 1 - x^2 - xh - x + xh -1\).Simplified numerator: \( -2h \).Simplify:\(\frac{-2h}{h(x+h-1)(x-1)} = \frac{-2}{(x+h-1)(x-1)}\).Take the limit as \(h\to 0\):\(\lim_{h \to 0} \frac{-2}{(x+h-1)(x-1)} = \frac{-2}{(x-1)^2}\).Thus, \(\frac{d y}{d x} = \frac{-2}{(x-1)^2}\).
04

Derivative of a Quadratic Function

For the function \(y = 3x^2\), find \(f(x+h)\):\(f(x+h) = 3(x+h)^2 = 3(x^2 + 2xh + h^2) = 3x^2 + 6xh + 3h^2\).Difference quotient:\(\frac{f(x+h) - f(x)}{h} = \frac{(3x^2 + 6xh + 3h^2) - 3x^2}{h} = \frac{6xh + 3h^2}{h}\).Simplify:\(6x + 3h\).Take the limit as \(h\to 0\): \(\lim_{h\to 0} (6x + 3h) = 6x\).Therefore, \(\frac{d y}{d x} = 6x\).

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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 fundamental concept in calculus. It is used to estimate the slope of the tangent line to a curve at a point. This slope is essentially the derivative of a function. For a function given by \(f(x)\), the difference quotient is expressed as:
  • \( \frac{f(x+h) - f(x)}{h} \)
This formula involves evaluating the function at two close points, \(x\) and \(x + h\), and measuring how much the function's value changes relative to \(h\). Once simplified, we take the limit as \(h\) approaches zero to find the exact derivative of the function. The entire process captures how the function behaves infinitely close to a certain point, providing key insights into its geometric behavior.
Limit Definition of Derivative
To find the derivative from first principles, we use the limit definition of a derivative. This involves calculating the limit of the difference quotient as \(h\) goes to zero. Mathematically, the derivative \(f'(x)\) is given by:
  • \( \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)
This limit determines how the function changes at an exact point. The key idea is that, as \(h\) becomes infinitesimally small, the difference quotient closely approximates the slope of the tangent to the curve at that particular point. This definition underpins the fundamental calculus concept of derivative and paves the way for determining rates of change in many contexts.
Linear Functions Differentiation
Linear functions are the simplest to differentiate. Consider a linear function like \(y = 6 - 7x\). To find its derivative using first principles:
  1. Compute \(f(x+h)\): becomes \(6 - 7(x+h)\) simplifying to \(6 - 7x - 7h\).
  2. Apply difference quotient: \( \frac{(6 - 7x - 7h) - (6 - 7x)}{h} = \frac{-7h}{h}\).
  3. Simplify to get: \(-7\).
  4. Applying the limit: As \(h\) approaches zero, the expression stays at \(-7\). The derivative of \(y\) with respect to \(x\) is thus \(-7\).
This illustrates that the slope of a linear function is constant, as expected, corroborating that its graph is a straight line with a consistent gradient.
Rational Functions Differentiation
For rational functions, such as \(y = \frac{x+1}{x-1}\), finding derivatives involves more detailed algebraic manipulation. Here鈥檚 the process:
  1. Compute \(f(x+h)\): Simplifies to \(\frac{x+h+1}{x+h-1}\).
  2. Difference quotient: \(\frac{\frac{x+h+1}{x+h-1} - \frac{x+1}{x-1}}{h}\).
  3. Combine using common denominators, simplify: This involves expanding polynomials and simplifying.
  4. Simplified to \(\frac{-2}{(x+h-1)(x-1)}\).
  5. Take limit as \(h\to 0\): Results in \(-\frac{2}{(x-1)^2}\).
Rational functions often require meticulous fraction manipulation. The derivative reveals insights, like potential asymptotic behaviors and discontinuities of the graph.
Quadratic Functions Differentiation
Quadratic functions, like \(y = 3x^2\), involve polynomial expressions, and here's how to differentiate them from first principles:
  1. Calculate \(f(x+h)\): Expanding \((x+h)^2\) gives \(x^2 + 2xh + h^2\), leading to \(3x^2 + 6xh + 3h^2\).
  2. Apply the difference quotient: \(\frac{3x^2 + 6xh + 3h^2 - 3x^2}{h} = \frac{6xh + 3h^2}{h}\).
  3. Simplify: Results in \(6x + 3h\).
  4. Take the limit as \(h\to 0\): Simplifies directly to \(6x\).
This demonstrates that for quadratics, the derivative gives the slope of the tangent at any point \(x\). It shows how the curve's steepness changes linearly, consistent with the linear nature of the derived formula.

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Most popular questions from this chapter

Determine the equation of the tangent to the curve \(y=\left(x^{3}-5 x+2\right)\left(3 x^{2}-2 x\right)\) at the point (1,-2)

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