Chapter 2: Problem 12
For each function, use the definition of the derivative to determine \(\frac{d y}{d x},\) where \(a, b, c,\) and \(m\) are constants. a. \(y=c\) b. \(y=x\) c. \(y=m x+b\) d. \(y=a x^{2}+b x+c\)
Short Answer
Expert verified
\( \frac{dy}{dx} \) for (a) 0, (b) 1, (c) \( m \), (d) \( 2ax+b \).
Step by step solution
01
Understanding the Concept of Derivative
The derivative of a function represents the rate at which the function's value changes as the input changes. Mathematically, the derivative of a function \( y=f(x) \) at a point \( x \) is defined as: \( \frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \). We'll use this definition to find the derivative for each given function.
02
Derivative of a Constant Function (\(y = c\))
For a constant function \( y = c \), \( f(x) = c \) and thus \( f(x+h) = c \). The derivative is calculated as: \[ \frac{dy}{dx} = \lim_{h \to 0} \frac{c - c}{h} = \lim_{h \to 0} \frac{0}{h} = 0 \] So, the derivative of a constant is zero.
03
Derivative of the Linear Function (\(y = x\))
For the linear function \( y = x \), \( f(x) = x \) and \( f(x+h) = x+h \). The derivative is: \[ \frac{dy}{dx} = \lim_{h \to 0} \frac{(x + h) - x}{h} = \lim_{h \to 0} \frac{h}{h} = \lim_{h \to 0} 1 = 1 \] Thus, the derivative of \( y = x \) is 1.
04
Derivative of General Linear Function (\(y = mx + b\))
For \( y = mx + b \), \( f(x) = mx + b \) and \( f(x+h) = m(x+h) + b \). Calculate the derivative:\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{(m(x+h) + b) - (mx + b)}{h} = \lim_{h \to 0} \frac{mx + mh + b - mx - b}{h} \]\[ = \lim_{h \to 0} \frac{mh}{h} = \lim_{h \to 0} m = m \] So, the derivative of a linear equation \( y = mx + b \) is \( m \).
05
Derivative of a Quadratic Function (\(y = ax^2 + bx + c\))
For the quadratic function \( y = ax^2 + bx + c \), expand \( f(x+h) = a(x+h)^2 + b(x+h) + c \):\[ a(x+h)^2 = a(x^2 + 2xh + h^2) = ax^2 + 2axh + ah^2 \]\[ b(x+h) = bx + bh \]Thus, \( f(x+h) = ax^2 + 2axh + ah^2 + bx + bh + c \). Now, calculate the derivative:\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{(ax^2 + 2axh + ah^2 + bx + bh + c) - (ax^2 + bx + c)}{h} \]\[ = \lim_{h \to 0} \frac{2axh + ah^2 + bh}{h} = \lim_{h \to 0} (2ax + ah + b) \]As \( h \to 0 \), the derivative is \( 2ax + b \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
derivative
The derivative is a core concept in calculus and it's all about understanding how a function changes. Imagine a car on a road: the derivative is like the speedometer, showing how fast the car's position (function value) is changing with respect to time (input value). By taking a derivative, you essentially determine how one variable's change impacts the value of another. In mathematical terms, the derivative of a function captures this rate of change at any point. For example, if you have a function \( y = f(x) \), its derivative \( \frac{dy}{dx} \) is symbolically expressed using:
- The limit definition: \( \frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)
limit definition of derivative
In calculus, the limit definition of the derivative provides a rigorous mathematical way to describe how a function changes. Imagine describing a path you’re walking on: the derivative tells you if you’re going uphill, downhill, or moving flat.
- The formula \( \frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \) expresses this idea precisely.
- Here, \( h \) represents a tiny difference in the \( x \)-value.
- \( f(x+h) - f(x) \) is the corresponding tiny difference in the \( y \) value.
linear functions
Linear functions are the simplest forms of functions in mathematics that depict a straight line on a graph. These functions can be written in the form of \( y = mx + b \). Here, \( m \) is the slope, indicating how much \( y \) changes with \( x \), and \( b \) is the y-intercept, the point where the line crosses the y-axis.
- In terms of derivatives, calculating the derivative of a linear function is straightforward: it's simply the slope \( m \) of the line.
- For instance, considering the function \( y = x \), the derivative \( \frac{dy}{dx} = 1 \) because \( m = 1 \).
quadratic functions
Quadratic functions are slightly more complex than linear functions and are represented as \( y = ax^2 + bx + c \). This form creates a parabolic curve on a graph, which can open either upwards or downwards depending on the coefficient \( a \). The derivative of a quadratic function helps to understand how quickly the slope of the curve is changing at any point.
- The derivative of a quadratic function \( y = ax^2 + bx + c \) is \( \frac{dy}{dx} = 2ax + b \).
- This result shows that the rate of change in a quadratic function is linear in itself.