Derivative of `1/x`
- Derivative of `1/x`
- What is the Derivative of `1/x`?
- Proof of Derivative of `frac{1}{x}` by First Principle
- Real-life Application of Derivative of `frac{1}{x}`
- Solved Examples
- Practice Problems
- Frequently Asked Questions
Introduction
The derivative of the function \(f(x) = \frac{1}{x}\) is a fundamental concept in calculus that captures how the function changes at any given point on its domain. This function \(f(x) = \frac{1}{x}\) is called a reciprocal function. It can also be expressed as \(x^{-1}\). This function is particularly interesting because it is defined for all real numbers \(x\), except for \(x = 0\), where it is undefined due to division by zero. The graph of \(f(x) = \frac{1}{x}\) is a hyperbola which is symmetric about the origin.
What is the Derivative of `1/x`?
To differentiate `frac{1}{x}` using calculus, the function \(f(x) = \frac{1}{x}\) can be written as \(f(x) = x^{-1}\).
Now, applying the power rule, which states that if \(f(x) = x^n\), then \(f'(x) = n \cdot x^{n-1}\), we have:
\(f'(x) = -1 \cdot x^{-1 - 1} = -1 \cdot x^{-2} = -\frac{1}{x^2}\)
So, the derivative of \(f(x) = \frac{1}{x}\) is \(f'(x) = -\frac{1}{x^2}\).
The derivative of \(f(x) = \frac{1}{x}\) describes the rate of change of the function with respect to its input \(x\). In theoretical terms, it quantifies how the function's value changes as \(x\) varies. For \(f(x) = \frac{1}{x}\), the derivative is \(f'(x) = -\frac{1}{x^2}\).
This concept is fundamental in understanding the behavior of functions and is widely applicable across various fields of mathematics and science.
Proof of Derivative of `frac{1}{x}` by First Principle
The first principle of derivatives, also known as the limit definition of the derivative, can be used to prove the derivative of \(f(x) = \frac{1}{x}\). According to this principle, the derivative of a function \(f(x)\) at a point \(x\) is given by:
`f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}`
Let's apply this to \(f(x) = \frac{1}{x}\).
`f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}`
To simplify the expression inside the limit, find a common denominator for the fractions in the numerator:
`f'(x) = \lim_{h \to 0} \frac{\frac{x - (x + h)}{x(x + h)}}{h}`
`f'(x) = \lim_{h \to 0} \frac{\frac{-h}{x(x + h)}}{h}`
`f'(x) = \lim_{h \to 0} \frac{-h}{h \cdot x(x + h)}`
Cancel \(h\) in the numerator and the denominator:
`f'(x) = \lim_{h \to 0} \frac{-1}{x(x + h)}`
Now, apply the limit \(h \to 0\):
\(f'(x) = \frac{-1}{x(x + 0)}\)
\(f'(x) = \frac{-1}{x^2}\)
Hence, proved that the derivative of \(f(x) = \frac{1}{x}\) is \(f'(x) = -\frac{1}{x^2}\).
Real-life Applications of Derivative of `\frac{1}{x}`
The derivative of `\frac{1}{x}` finds applications in various fields, notably in physics, engineering, and finance. One significant real-life application is in modeling rates of change, such as:
In Physics: In physics, this derivative is used to determine the strength of gravitational fields and electrical fields, where the force weakens with increasing distance according to an inverse square law. For example,
- The intensity of light, `I`, decreases with the distance from a light source, `d`. However, instead of a simple inverse proportion between `I` and `d`, `I` is inversely proportional to `d^2`.
- The gravitational pull `F` is inversely proportional to `r^2`, where `r` is the distance between the centers of the two masses.
In Finance: In finance, it's utilized in the calculation of marginal utility, which measures the additional satisfaction gained from consuming an additional unit of a good. It's also employed in analyzing interest rates and depreciation of assets.
In Engineering: Engineers use it in fields like fluid dynamics to describe flow rates, pressure differentials, and other variables that vary inversely with distance or size.
Understanding the derivative of `\frac{1}{x}` helps in comprehending how quantities change with respect to each other and is essential for optimizing processes and systems in various real-world scenarios.
Solved Examples
Example `1`. Finding the derivative of \( \frac{1}{2x} \).
Solution:
Using the power rule for differentiation:
\( f(x) = \frac{1}{2x} = \frac{1}{2}x^{-1} \)
Now, applying the power rule:
\( f'(x) = -1 \cdot \frac{1}{2}x^{-1-1} = -\frac{1}{2}x^{-2} = -\frac{1}{2x^2} \)
The derivative of \( f(x) = \frac{1}{2x} \) is \( f'(x) = -\frac{1}{2x^2} \).
Example `2`. For \( f(x) = \frac{4}{3x} \), find \( f'(x) \), the derivative of \( f(x) \).
Solution:
Using the power rule for differentiation:
\( f(x) = \frac{4}{3x} = \frac{4}{3} \cdot \frac{1}{x} = \frac{4}{3} \cdot x^{-1} \)
Now, applying the power rule:
\( f'(x) = \frac{4}{3} (-1 \cdot x^{-1-1}) = -\frac{4}{3}x^{-2} = -\frac{4}{3x^2} \)
The derivative of \( f(x) = \frac{4}{3x} \) is \( f'(x) = -\frac{4}{3x^2} \).
Example `3`. Use the First Principle to differentiate \( f(x) = \frac{-2}{6x} \) with respect to `x`.
Solution:
We'll use the limit definition of the derivative:
`f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}`
First, let's express \( f(x + h) \) and \( f(x) \):
\( f(x + h) = \frac{-2}{6(x + h)} \)
\( f(x) = \frac{-2}{6x} \)
Substitute these into the limit definition:
`f'(x) = \lim_{h \to 0} \frac{\frac{-2}{6(x + h)} - \frac{-2}{6x}}{h}`
`= \lim_{h \to 0} \frac{-2(6x) + 2(6(x + h))}{6x(6(x + h))h}`
`= \lim_{h \to 0} \frac{-12x + 12(x + h)}{36x(x + h)h}`
`= \lim_{h \to 0} \frac{-12x + 12x + 12h}{36x(x + h)h}`
`= \lim_{h \to 0} \frac{12h}{36x(x + h)h}`
`= \lim_{h \to 0} \frac{12}{36x(x + h)}`
`= \frac{12}{36x^2}`
`= \frac{1}{3x^2}`
The derivative of \( f(x) = \frac{-2}{6x} \) using the first principle is \( f'(x) = \frac{1}{3x^2} \).
Example `4`: Find the derivative of \( f(x) = \frac{2}{x + 1} \).
Solution:
Let's find the derivative of \( f(x) = \frac{2}{x + 1} \) using the power rule.
The power rule states that if we have a function of the form \( f(x) = ax^n \), then its derivative is given by:
\( f'(x) = n \cdot ax^{n-1} \)
In our case, \( f(x) = \frac{2}{x + 1} \), which can be rewritten as \( f(x) = 2(x + 1)^{-1} \).
Now, using the power rule, we differentiate:
\( f'(x) = -1 \cdot 2(x + 1)^{-1-1} \)
\( f'(x) = -2(x + 1)^{-2} \)
\( f'(x) = -\frac{2}{(x + 1)^2} \)
So, the derivative of \( f(x) = \frac{2}{x + 1} \) using the power rule is \( f'(x) = -\frac{2}{(x + 1)^2} \).
Example `5`. Find the derivative of \( f(x) = \frac{5}{3x^2 + 7} \).
Solution:
Let's find the derivative of the function \( f(x) = \frac{5}{3x^2 + 7} \).
We can rewrite this function as \( f(x) = 5(3x^2 + 7)^{-1} \).
Now, using the power rule and the chain rule, we differentiate:
\( f'(x) = -1 \cdot 5 \cdot (3x^2 + 7)^{-1-1} \cdot (6x) \)
\( f'(x) = -\frac{30x}{(3x^2 + 7)^2} \)
So, the derivative of \( f(x) = \frac{5}{3x^2 + 7} \) using the power rule is \( f'(x) = -\frac{30x}{(3x^2 + 7)^2} \).
Practice Problems
Q`1`. Find the derivative of the function \( f(x) = -\frac{1}{x} \).
- \( \frac{1}{x^2} \)
- \( -\frac{1}{x^3} \)
- \( -\frac{1}{x^2} \)
- \( -\frac{3}{x^3} \)
Solution: a
Q`2`. Differentiate \( g(x) = \frac{1}{3x} \).
- \( -\frac{1}{3x^3} \)
- \( -\frac{1}{3x^2} \)
- \( -\frac{1}{4x^2} \)
- \( -\frac{3}{3x^2} \)
Solution: b
Q`3`. Calculate the derivative of \( h(x) = \frac{2}{x^3} \).
- \( -\frac{3}{x^2} \)
- \( -\frac{5}{x^4} \)
- \( -\frac{6}{x^4} \)
- \( -\frac{6}{x^2} \)
Solution: c
Q`4`. Find the derivative of \( f(x) = \frac{5}{x^2} \).
- \( -\frac{7}{x^3} \)
- \( -\frac{10}{x^2} \)
- \( -\frac{5}{x^3} \)
- \( -\frac{10}{x^3} \)
Solution: d
Q`5`. Differentiate \( g(x) = \frac{1}{2x^4} \).
- \( -\frac{2}{x^5} \)
- \( -\frac{2}{x^4} \)
- \( -\frac{4}{x^5} \)
- \( -\frac{1}{x^5} \)
Solution: a
Q`6`. Evaluate the derivative of \( h(x) = \frac{3}{\sqrt{x}} \).
- \( -\frac{2}{3\sqrt{x^2}} \)
- \( -\frac{5}{2\sqrt{x}} \)
- \( -\frac{3}{2\sqrt{x^3}} \)
- \( -\frac{1}{2\sqrt{x}} \)
Solution: c
Frequently Asked Questions
Q`1`. What is the derivative of \( \frac{1}{x} \)?
Answer: The derivative of \( \frac{1}{x} \) with respect to \( x \) is \( -\frac{1}{x^2} \).
Q`2`. Why is the derivative of \( \frac{1}{x} \) negative?
Answer: The derivative of \( \frac{1}{x} \) is \( -\frac{1}{x^2} \). The derivative is always negative because `x^2 > 0` which implies `1/(x^2) > 0`. Hence `-1/(x^2) < 0`.
Q`3`. What is the geometric interpretation of the derivative of \( \frac{1}{x} \)?
Answer: Geometrically, the derivative of \( \frac{1}{x} \) represents the slope of the tangent line to the curve \( y = \frac{1}{x} \) at any given point \( (x, \frac{1}{x}) \) on the curve.
Q`4`. Where is the derivative of \( \frac{1}{x} \) undefined?
Answer: The derivative is undefined at \( x = 0 \) because \( \frac{1}{x} \) is not differentiable at \( x = 0 \) as dividing by zero makes any fraction undefined.
Q`5`. What are some applications of the derivative of \( \frac{1}{x} \)?
Answer: The derivative of \( \frac{1}{x} \) is used in various fields such as physics, economics, and engineering to analyze rates of change, growth, and decay phenomena.