Problem 26
Question
Approximate \(f^{\prime}(1)\) by considering the difference quotient $$ \frac{f(1+h)-f(1)}{h} $$ for values of \(h\) near \(0,\) and then find the exact value of \(f^{\prime}(1)\) by differentiating. $$ f(x)=\frac{1}{x^{2}} $$
Step-by-Step Solution
Verified Answer
The exact value of \( f^{\prime}(1) \) is \( -2 \).
1Step 1: Write the difference quotient
The difference quotient given in the problem is \( \frac{f(1+h) - f(1)}{h} \). We start by determining \( f(1) \) and \( f(1+h) \). For the function \( f(x) = \frac{1}{x^2} \), we find \( f(1) = \frac{1}{1^2} = 1 \) and \( f(1+h) = \frac{1}{(1+h)^2} \). Thus, the difference quotient becomes \( \frac{\frac{1}{(1+h)^2} - 1}{h} \).
2Step 2: Simplify the difference quotient
Substitute \( f(1) \) and \( f(1+h) \) into the quotient: \[\frac{\frac{1}{(1+h)^2} - 1}{h} = \frac{\frac{1 - (1+h)^2}{(1+h)^2}}{h} = \frac{1 - (1 + 2h + h^2)}{h(1+h)^2} = \frac{-2h - h^2}{h(1+h)^2}.\]Divide numerator and denominator by \( h \): \[\frac{-2 - h}{(1+h)^2}.\]
3Step 3: Approximate by letting \( h \to 0 \)
As \( h \) approaches 0, the expression \( \frac{-2 - h}{(1+h)^2} \) simplifies by substituting \( h = 0 \), which gives us \( \frac{-2}{1^2} = -2 \). Thus, \( f^{\prime}(1) \approx -2 \) using the difference quotient.
4Step 4: Find the derivative
To find the exact value of \( f'(x) \), differentiate the function \( f(x) = \frac{1}{x^2} \) with respect to \( x \). We can use the power rule: rewrite \( f(x) \) as \( x^{-2} \) and differentiate to obtain \( f'(x) = -2x^{-3} \).
5Step 5: Evaluate the derivative at \( x = 1 \)
Plug \( x = 1 \) into the derivative \( f'(x) = -2x^{-3} \), yielding \( f^{\prime}(1) = -2(1)^{-3} = -2 \). Thus, the exact value of \( f^{\prime}(1) \) is \( -2 \).
Key Concepts
Difference QuotientDifferentiationPower Rule
Difference Quotient
The difference quotient is a central concept in calculus, often used to find the derivative of a function at a specific point. Think of it like a tool for measuring how a function changes, or its "slope," at any given spot.
For the function \( f(x) = \frac{1}{x^2} \), the difference quotient at \( x = 1 \) is given by the formula:
To use this, we first find \( f(1) = 1 \) and \( f(1+h) = \frac{1}{(1+h)^2} \). Plugging these into the difference quotient gives:
For the function \( f(x) = \frac{1}{x^2} \), the difference quotient at \( x = 1 \) is given by the formula:
- \( \frac{f(1+h)-f(1)}{h} \)
To use this, we first find \( f(1) = 1 \) and \( f(1+h) = \frac{1}{(1+h)^2} \). Plugging these into the difference quotient gives:
- \( \frac{\frac{1}{(1+h)^2} - 1}{h} \)
Differentiation
Differentiation is the process of finding a derivative, which tells us the rate at which a function is changing. It's like asking how fast something is happening at a particular instance.
For the function \( f(x) = \frac{1}{x^2} \), we first rewrite it in a form that's easier to differentiate: \( x^{-2} \). This transformation makes it simpler to apply rules of differentiation.
Using the power rule, which you'll learn more about shortly, we differentiate to find \( f'(x) = -2x^{-3} \). This new function, \( f'(x) \), reveals how the original function \( f(x) \) changes with respect to \( x \).
At \( x = 1 \), plug into the differentiated formula: \( f^{\prime}(1) = -2(1)^{-3} \). And we find that \( f^{\prime}(1) = -2 \), confirming the behavior indicated by our difference quotient: that the function's "slope" or rate of change at \( x = 1 \) is -2.
For the function \( f(x) = \frac{1}{x^2} \), we first rewrite it in a form that's easier to differentiate: \( x^{-2} \). This transformation makes it simpler to apply rules of differentiation.
Using the power rule, which you'll learn more about shortly, we differentiate to find \( f'(x) = -2x^{-3} \). This new function, \( f'(x) \), reveals how the original function \( f(x) \) changes with respect to \( x \).
At \( x = 1 \), plug into the differentiated formula: \( f^{\prime}(1) = -2(1)^{-3} \). And we find that \( f^{\prime}(1) = -2 \), confirming the behavior indicated by our difference quotient: that the function's "slope" or rate of change at \( x = 1 \) is -2.
Power Rule
The power rule is one of the most fundamental techniques in calculus for differentiation. If you have a function in the form \( x^n \), its derivative is \( nx^{n-1} \).
This rule simplifies the process of differentiation, making it quick and efficient for polynomial functions. When applying the power rule to our function \( f(x) = x^{-2} \), you multiply the power by the coefficient (which is 1 in this case), and decrease the power by one:
By practicing with a simple function like \( x^n \), you can build a strong foundation for tackling more complex problems using differentiation techniques.
This rule simplifies the process of differentiation, making it quick and efficient for polynomial functions. When applying the power rule to our function \( f(x) = x^{-2} \), you multiply the power by the coefficient (which is 1 in this case), and decrease the power by one:
- The derivative is \( -2x^{-2-1} \)
- This simplifies to \( -2x^{-3} \)
By practicing with a simple function like \( x^n \), you can build a strong foundation for tackling more complex problems using differentiation techniques.
Other exercises in this chapter
Problem 26
Find \(g^{\prime}(3)\) given that \(f(3)=-2\) and \(f^{\prime}(3)=4\). (a) \(g(x)=3 x^{2}-5 f(x)\) (b) \(g(x)=\frac{2 x+1}{f(x)}\)
View solution Problem 26
Find the equation of the line tangent to the graph of \(\sin x\) at $$ \text { (a) } x=0 \quad \text { (b) } x=\pi \quad \text { (c) } x=\pi / 4 $$
View solution Problem 27
Find \(d y / d x\) $$ y=x^{3} \sin ^{2}(5 x) $$
View solution Problem 27
In parts \((\mathrm{a})-(\mathrm{d}), F(x)\) is expressed in terms of \(f(x)\) and \(g(x) .\) Find \(F^{\prime}(2)\) given that \(f(2)=-1, f^{\prime}(2)=4, g(2)
View solution