Problem 26
Question
Use the formal definition to find the derivative of \(f(x)=\frac{1}{x+1}\) at \(x=0\).
Step-by-Step Solution
Verified Answer
The derivative at \( x=0 \) is \( f'(0) = -1 \).
1Step 1: Understand the Derivative Definition
The derivative of a function at a point, using the formal definition, is given by the limit formula: \[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]Here, we want to find \( f'(0) \). So, \( a = 0 \) and our function is \( f(x) = \frac{1}{x+1} \).
2Step 2: Substitute into the Derivative Formula
Apply the function \( f(x) = \frac{1}{x+1} \) into the derivative definition:\[ f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} \]This becomes:\[ f'(0) = \lim_{h \to 0} \frac{\frac{1}{h+1} - \frac{1}{1}}{h} \]
3Step 3: Simplify the Expression
Simplify the expression inside the limit:\[ \frac{\frac{1}{h+1} - 1}{h} = \frac{\frac{1 - (h+1)}{h+1}}{h} = \frac{\frac{-h}{h+1}}{h} \]This simplifies further to:\[ \frac{-h}{h(h+1)} = \frac{-1}{h+1} \]
4Step 4: Compute the Limit
Now, compute the limit as \( h \) approaches 0:\[ f'(0) = \lim_{h \to 0} \frac{-1}{h+1} = \frac{-1}{1} = -1 \]
5Step 5: State the Result
The derivative of \( f(x) = \frac{1}{x+1} \) at \( x = 0 \) is \( f'(0) = -1 \).
Key Concepts
DerivativeFormal Definition of DerivativeLimit Formula
Derivative
The concept of a derivative is fundamental in calculus. It represents the rate at which a function's value changes at any given point.
Imagine you are driving, and the speedometer gives you your speed at every instance, your speed is the derivative of your travel distance with respect to time.
Mathematically, if we have a function, say \( f(x) \), the derivative of that function, denoted as \( f'(x) \) or \( \frac{df}{dx} \), tells us how \( f(x) \) changes as \( x \) changes.
Imagine you are driving, and the speedometer gives you your speed at every instance, your speed is the derivative of your travel distance with respect to time.
Mathematically, if we have a function, say \( f(x) \), the derivative of that function, denoted as \( f'(x) \) or \( \frac{df}{dx} \), tells us how \( f(x) \) changes as \( x \) changes.
- A positive derivative means the function is increasing.
- A negative derivative means the function is decreasing.
- If the derivative is zero, the function might have reached a peak or a flat point.
Formal Definition of Derivative
The formal definition of a derivative is a cornerstone in the study of calculus and might seem complex at first, but it simplifies the concept of 'change.'
The derivative at a point \( a \) for a function \( f(x) \) is defined using a limit:\[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]
This expression might look daunting, but here’s what it means:
The derivative at a point \( a \) for a function \( f(x) \) is defined using a limit:\[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]
This expression might look daunting, but here’s what it means:
- "\( h \)" is a tiny increment added to \( x \). Think of it as a small step.
- "\( f(a + h) - f(a) \)" is the change in the function’s value from \( a \) to \( a + h \).
- Division by "\( h \)" gives the rate of change, like the slope of a line.
- The limit "\( h \rightarrow 0 \)" ensures we look at an infinitesimally small change, capturing the function's instantaneous rate of change rather than an average.
Limit Formula
In calculus, the concept of a limit is pivotal. It allows mathematicians to handle notions of infinity and infinitesimally small quantities.
The limit formula used in the derivative's definition helps formalize how a function's behavior approaches a particular value.
In simple terms, the limit \( \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \) tells us what happens as \( h \) becomes very close to zero.
The limit formula used in the derivative's definition helps formalize how a function's behavior approaches a particular value.
In simple terms, the limit \( \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \) tells us what happens as \( h \) becomes very close to zero.
- It finds the output when the input is infinitesimally tiny.
- This helps smooth out 'jumps' and 'gaps' in understanding the function’s behavior.
- The use of limits is what ensures calculus can handle continuous change comfortably.
Other exercises in this chapter
Problem 25
In Problems 1-28, differentiate the functions with respect to the independent variable. $$ g(s)=\left(3 s^{7}-7 s\right)^{3 / 2} $$
View solution Problem 26
Calculate the linear approximation for \(f(x)\) : $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ . \(f(x)=e^{2 x+1}\) at \(a=-1 / 2\)
View solution Problem 26
Find the derivative with respect to the independent variable. $$ f(x)=-\cos \left(3 x^{3}-4 x\right) $$
View solution Problem 26
Differentiate the functions with respect to the independent variable. (Note that log denotes the logarithm to base 10.) $$ f(x)=\ln (4-3 x) $$
View solution