Chapter 2: Problem 22
Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative. $$g(t)=\frac{1}{\sqrt{t}}$$
Short Answer
Expert verified
The derivative is \( g'(t) = \frac{-1}{2t^{3/2}} \) with a domain of \( t > 0 \).
Step by step solution
01
Definition of the Derivative
The derivative of a function at a point is defined as the limit \( f'(t) = \lim_{h \to 0} \frac{f(t+h) - f(t)}{h} \). Here, our function is \( g(t) = \frac{1}{\sqrt{t}} \). We will apply this definition to find \( g'(t) \).
02
Set Up the Difference Quotient
First, compute the difference \( g(t+h) - g(t) \). For \( g(t) = \frac{1}{\sqrt{t}} \), we have \( g(t+h) = \frac{1}{\sqrt{t+h}} \). Therefore, the difference is \( \frac{1}{\sqrt{t+h}} - \frac{1}{\sqrt{t}} \).
03
Simplify the Difference Quotient
To simplify \( \frac{1}{\sqrt{t+h}} - \frac{1}{\sqrt{t}} \), find a common denominator, resulting in \( \frac{\sqrt{t} - \sqrt{t+h}}{\sqrt{t(t+h)}} \). Simplify further by multiplying the numerator and the denominator by the conjugate \( \sqrt{t} + \sqrt{t+h} \), yielding \( \frac{-(t+h) + t}{h(\sqrt{t(t+h)})(\sqrt{t} + \sqrt{t+h})} \).
04
Evaluate the Limit
Simplify the expression to \( \lim_{h \to 0} \frac{-h}{h(\sqrt{t(t+h)})(\sqrt{t} + \sqrt{t+h})} \). Cancel the \( h \) terms to get \( \lim_{h \to 0} \frac{-1}{\sqrt{t(t+h)}(\sqrt{t} + \sqrt{t+h})} \). As \( h \to 0 \), this limit simplifies to \( \frac{-1}{2t\sqrt{t}} = \frac{-1}{2t^{3/2}} \). Thus, \( g'(t) = \frac{-1}{2t^{3/2}} \).
05
Domain of the Original Function
The function \( g(t) = \frac{1}{\sqrt{t}} \) is only defined for \( t > 0 \) because the square root requires positive t-values.
06
Domain of the Derivative
Similarly, \( g'(t) = \frac{-1}{2t^{3/2}} \) also requires \( t > 0 \) because the denominator contains \( t^{3/2} \), which implies that \( t \) must be positive.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit of a Function
The limit of a function is a fundamental concept in calculus. It describes the behavior of a function as its input approaches a particular value. When dealing with derivatives, limits help us understand how a function changes at precise points.
To find the derivative of a function using its definition, we use the formula:
To find the derivative of a function using its definition, we use the formula:
- \( f'(t) = \lim_{h \to 0} \frac{f(t+h) - f(t)}{h} \)
This formula represents the slope of the tangent line to the function at any point \( t \).
Simplifying Expressions
Simplifying expressions is crucial for evaluating limits, especially when the initial form is complex or indeterminate.
Let's consider simplifying the expression \( \frac{1}{\sqrt{t+h}} - \frac{1}{\sqrt{t}} \). This step involves
Let's consider simplifying the expression \( \frac{1}{\sqrt{t+h}} - \frac{1}{\sqrt{t}} \). This step involves
- First finding a common denominator, resulting in \( \frac{\sqrt{t} - \sqrt{t+h}}{\sqrt{t(t+h)}} \).
- Then, multiply the numerator and denominator by the conjugate \( \sqrt{t} + \sqrt{t+h} \). This trick helps eliminate radicals which simplifies the work in limits.
Function Domain
Function domain is about finding all the possible input values (or x-values) for which a function is defined. For certain functions, the domain might be restricted due to operations like division by zero or taking the square root of a negative number.
In our function \( g(t) = \frac{1}{\sqrt{t}} \), the square root limits the domain to values greater than zero \( t > 0 \) as the square root of a negative number is not defined in the real numbers.
Similarly, the domain of its derivative \( g'(t) = \frac{-1}{2t^{3/2}} \) also requires \( t > 0 \). This is because the term \( t^{3/2} \) in the denominator must remain positive to avoid division by zero or trying to take a square root of a negative.
In our function \( g(t) = \frac{1}{\sqrt{t}} \), the square root limits the domain to values greater than zero \( t > 0 \) as the square root of a negative number is not defined in the real numbers.
Similarly, the domain of its derivative \( g'(t) = \frac{-1}{2t^{3/2}} \) also requires \( t > 0 \). This is because the term \( t^{3/2} \) in the denominator must remain positive to avoid division by zero or trying to take a square root of a negative.
Differentiation Techniques
Differentiation techniques present various methods to find the derivative of a function. While using the definition of a derivative is foundational, other techniques often streamline the process.
In this particular exercise, we focus on utilizing the limit definition. This approach involves:
In this particular exercise, we focus on utilizing the limit definition. This approach involves:
- Setting up and simplifying a difference quotient after substituting function values
- Applying algebraic tricks like using conjugates to simplify expressions further
- Evaluating the resulting limit as \( h \to 0 \)