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Magazine Chapter 3: Problem 3
Using the definition, calculate the derivatives of the functions in Exercises \(1-6 .\) Then find the values of the derivatives as specified. $$ g(t)=\frac{1}{t^{2}} ; \quad g^{\prime}(-1), g^{\prime}(2), g^{\prime}(\sqrt{3}) $$
Short Answer
Expert verified
The derivatives are: \( g'(-1) = 2 \), \( g'(2) = -\frac{1}{4} \), and \( g'(\sqrt{3}) = -\frac{2\sqrt{3}}{9} \).
Step by step solution
01
Write Down the Function
The given function is \( g(t) = \frac{1}{t^2} \). This is the function for which we need to find the derivative.
02
Apply the Limit Definition of Derivative
The derivative of a function \( g(t) \) at a point \( a \) is given by \( g'(a) = \lim_{h \to 0} \frac{g(a+h) - g(a)}{h} \). Our task is to compute \( g'(t) \) using this definition.
03
Substitute the Function into the Derivative Formula
Substitute \( g(t) = \frac{1}{t^2} \) into the derivative formula:\[g'(t) = \lim_{h \to 0} \frac{\frac{1}{(t+h)^2} - \frac{1}{t^2}}{h}\]
04
Simplify the Expression
To simplify \( \frac{\frac{1}{(t+h)^2} - \frac{1}{t^2}}{h} \), first combine the fractions in the numerator:\[\frac{t^2 - (t+h)^2}{t^2(t+h)^2}\]
05
Simplify the Combined Fraction
Expand \((t+h)^2\) to get \(t^2 + 2th + h^2\). Subtract \( t^2 \) and simplify:\[g'(t) = \lim_{h \to 0} \frac{-2th - h^2}{h \cdot t^2(t+h)^2}\]
06
Cancel and Simplify Further
Factor \( h \) out of the numerator:\[g'(t) = \lim_{h \to 0} \frac{h(-2t-h)}{h \cdot t^2(t+h)^2} = \lim_{h \to 0} \frac{-2t-h}{t^2(t+h)^2}\]
07
Compute the Limit
As \( h \to 0 \), the expression simplifies to:\[g'(t) = \frac{-2t}{t^2 \cdot t^2} = \frac{-2}{t^3}\]
08
Evaluate the Derivative at Specific Points
Finally, we evaluate \( g'(t) \) at \( t = -1 \), \( t = 2 \), and \( t = \sqrt{3} \):- \( g'(-1) = \frac{-2}{(-1)^3} = 2 \)- \( g'(2) = \frac{-2}{2^3} = -\frac{1}{4} \)- \( g'(\sqrt{3}) = \frac{-2}{(\sqrt{3})^3} = -\frac{2}{3\sqrt{3}} = -\frac{2\sqrt{3}}{9} \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit Definition of Derivative
The limit definition of a derivative is a fundamental concept in calculus. It provides a precise way to find the rate of change or the slope of a function at any specific point. The definition states that the derivative of a function \( g(t) \) at a specific point \( a \) is given by:\[ g'(a) = \lim_{h \to 0} \frac{g(a+h) - g(a)}{h} \]This expression can be thought of as the difference quotient. Here:
- \( h \) is an incremental change in \( t \), approaching zero.
- \( g(a+h) - g(a) \) represents the change in the function's output.
- Dividing by \( h \) gives the average rate of change, and taking the limit as \( h \) approaches zero gives the instantaneous rate of change (the derivative).
Rational Functions
Rational functions are ratios of polynomials. They are written as \( \frac{P(t)}{Q(t)} \), where \( P(t) \) and \( Q(t) \) are polynomials with \( Q(t) eq 0 \). For this problem, the function \( g(t) = \frac{1}{t^2} \) is a rational function with:
- Numerator \( 1 \) and denominator \( t^2 \).
- The polynomial for \( Q(t) \) is \( t^2 \), which delineates undefined points where \( t = 0 \).
Derivative Evaluation
Evaluating a derivative involves substituting the original function into the limit definition and simplifying. For the given function \( g(t) = \frac{1}{t^2} \), the derivative using the limit definition is:\[ g'(t) = \lim_{h \to 0} \frac{\frac{1}{(t+h)^2} - \frac{1}{t^2}}{h} \]This requires simplifying complex fractions by obtaining a common denominator in the numerator expression:\[ \frac{t^2 - (t+h)^2}{t^2(t+h)^2} \]The next step is to expand and simplify the expression:
- Expand \((t+h)^2\) to \(t^2 + 2th + h^2\).
- Subtract \(t^2\) and factor out \(h\).
Derivative Simplification
Simplifying the derivative expression ensures the calculation is manageable, especially for rational functions. After finding:\[ g'(t) = \lim_{h \to 0} \frac{-2th - h^2}{h \cdot t^2(t+h)^2} \] Factor and cancel \( h \) in the numerator and denominator:\[ g'(t) = \lim_{h \to 0} \frac{-2t-h}{t^2(t+h)^2} \]As \( h \) approaches 0, the \(-h\) term vanishes, so the expression further simplifies, and you compute the derivative: \[ g'(t) = \frac{-2}{t^3} \]This result is then used to find the derivative at specific points like \( t = -1 \), \( t = 2 \), and \( t = \sqrt{3} \), demonstrating the power of derivative simplification.
