Derivative of `ln`
- Introduction
- What is the derivative of the natural `log` of `x`
- `ln` Rules
- Derivative of `ln` by First Principle
- Differentiate `ln(x)` by implicit differentiation
- Real-life application of Derivative of `ln`
- Solved Examples on Derivative of `ln`
- Practice Problems on Derivative of `ln`
- Frequently Asked Questions on Derivative of `ln`
Introduction
`ln x` is a natural logarithmic function. We can also call it a logarithm with base `e`. Derivative of `ln x` means that the rate of change of the natural logarithm of \(x\) with respect to \(x\). The domain of the function, `ln x`, are all positive real numbers, not including zero. The range of the function is all real numbers. The function is always increasing for all intervals.
What is the derivative of the natural `log` of `x`
The derivative of the natural logarithm function or derivative of `lnx`, denoted as \(\ln(x)\), is defined as follows:
This means that the rate of change of the natural logarithm of \(x\) with respect to \(x\) is equal to the reciprocal of \(x\). In other words, as \(x\) changes, the rate at which \(\ln(x)\) changes is inversely proportional to \(x\). The domain and range of the derivative of `ln x` is all positive real numbers excluding zero.
The derivative of \(\ln(x)\) is a fundamental result in calculus and is often used in various mathematical and scientific applications. It states that the rate of change of \(\ln(x)\) with respect to \(x\) is inversely proportional to \(x\).
Proof of Derivative of `ln x` by First Principle
To find the derivative of the natural logarithm function \(\ln(x)\) using the first principles (also known as the definition of the derivative), we start with the limit definition of the derivative:
\( f'(x) = \lim_{{h \to 0}} \frac{f(x + h) - f(x)}{h} \)
In this case, \(f(x) = \ln(x)\). Let's apply the definition to find the derivative:
\( \frac{d}{dx}(\ln(x)) = \lim_{{h \to 0}} \frac{\ln(x + h) - \ln(x)}{h} \)
Now, use the property of logarithms that \(\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)\):
\( = \lim_{{h \to 0}} \frac{\ln\left(\frac{x + h}{x}\right)}{h} \)
Combine the logarithmic fraction into a single logarithm:
\( = \lim_{{h \to 0}} \frac{\ln\left(1 + \frac{h}{x}\right)}{h} \)
Now, apply a substitution \(u = \frac{h}{x}\). As \(h\) approaches `0`, \(u\) approaches `0`.
\( = \lim_{{u \to 0}} \frac{\ln(1 + u)}{ux} \)
Applying the product property of limit:
\( = \lim_{{u \to 0}} \frac{\ln(1 + u)}{u}\lim_{{u \to 0}} \frac{1}{x} \)
As we know \( \lim_{{x \to 0}} \frac{\ln(1 + x)}{x} = 1 \), it becomes
\( = 1 \times \frac{1}{x} \)
\( = \frac{1}{x} \)
Therefore, by the first principles, the derivative of \(\ln(x)\) with respect to \(x\) is \(\frac{1}{x}\).
Proof of Derivative of `ln x` by implicit differentiation
To differentiate \(\ln(x)\) with respect to \(x\) using implicit differentiation, we will consider \(y = \ln(x)\)
let's first convert \( \ln(x) \) to its equivalent exponential form.
The logarithmic function \( \ln(x) \) can be expressed in exponential form as:
\( \ln(x) = y \)
\( \Rightarrow e^y = x \)
Now, let's differentiate both sides of the equation implicitly with respect to \( x \):
\( \frac{d}{dx}(e^y) = \frac{d}{dx}(x) \)
Now, we apply the chain rule on the left side:
\( \frac{d}{dx}(e^y) = \frac{d}{dy}(e^y) \cdot \frac{dy}{dx} \)
\( = e^y \cdot \frac{dy}{dx} \)
And on the right side, the derivative of \( x \) with respect to \( x \) is simply `1`.
So, our equation becomes:
\( e^y \cdot \frac{dy}{dx} = 1 \)
\( \frac{dy}{dx} = \frac{1}{e^y} \)
Since we had \( y = \ln(x) \), we can replace \( y \) with \( \ln(x) \):
\( \frac{dy}{dx} = \frac{1}{e^{\ln(x)}} \)
\( = \frac{1}{x} \)
Hence, the derivative of \( \ln(x) \) with respect to \( x \) is \( \frac{1}{x} \).
Real-life Applications of Derivative of `ln`
Rate of Growth or Decay: In fields like biology, economics, and physics, the derivative of `ln(x)` can represent the rate of growth or decay of a quantity over time. For example, in population growth models or radioactive decay processes, the rate at which something changes with respect to time can often be described using `ln(x)`.
Chemical Kinetics: In chemistry, the rate of reaction in many chemical processes can be described using the derivative of `ln(x)`. For instance, in first-order reactions, the rate at which a reactant is consumed or a product is formed often follows an exponential decay, and the derivative of `ln(x)` helps in quantifying this rate.
Finance and Economics: In finance and economics, the derivative of `ln(x)` often comes into play when analyzing interest rates, investment returns, and economic growth rates. For example, in continuous compounding interest calculations, the derivative of `ln(x)` helps determine the instantaneous rate of change of an investment's value over time.
Solved Examples
Example `1`. Find the derivative of the function \( f(x) = \ln(x^2 + 1) \).
Solution:
Given function: \( f(x) = \ln(x^2 + 1) \)
To find the derivative \( f'(x) \), we'll use the chain rule:
\( f'(x) = \frac{d}{dx} \ln(u) \)
\( = \frac{1}{u} \cdot \frac{du}{dx} \)
where \( u = x^2 + 1 \).
Now, we'll find \( \frac{du}{dx} \):
\( \frac{du}{dx} = \frac{d}{dx}(x^2 + 1) \)
\( = 2x \)
Substitute \( u = x^2 + 1 \) and \( \frac{du}{dx} = 2x\),
\( f'(x) = \frac{1}{x^2 + 1} \cdot 2x \)
\( = \frac{2x}{x^2 + 1} \)
Example `2`. Calculate the critical points of the function \( g(x) = x \ln(x) \) in the interval \( (0, \infty) \).
Solution:
Given function: \( g(x) = x \ln(x) \)
To find the critical points, we need to find where the derivative is equal to zero or undefined.
\( g'(x) = \frac{d}{dx}(x \ln(x)) \)
Using the product rule:
\( g'(x) = \ln(x) + x \cdot \frac{1}{x} \)
\( = \ln(x) + 1 \)
To find critical points, set the derivative equal to zero:
\( \ln(x) + 1 = 0 \)
\( \ln(x) = -1 \)
Exponentiating both sides:
\( e^{\ln(x)} = e^{-1} \)
\( x = e^{-1} = \frac{1}{e} \)
Since \( x \) must be positive, \( x = \frac{1}{e} \) is the only critical point in the interval \( (0, \infty) \).
Example `3`. Find the slope of the tangent on the curve \( f(x) = \ln(\sqrt{x} + 1) \) at `x = 4`.
Solution:
Given function: \( f(x) = \ln(\sqrt{x} + 1) \)
To find the derivative \( f'(x) \), we'll again use the chain rule:
\( f'(x) = \frac{d}{dx} \ln(u) \)
\( = \frac{1}{u} \cdot \frac{du}{dx} \)
where \( u = \sqrt{x} + 1 \).
Now, we'll find \( \frac{du}{dx} \):
\( \frac{du}{dx} = \frac{d}{dx}(\sqrt{x} + 1) \)
\( = \frac{1}{2\sqrt{x}} \)
Substitute \( u = \sqrt{x} + 1 \) and \( \frac{du}{dx} = \frac{1}{2\sqrt{x}}\),
\( f'(x) = \frac{1}{\sqrt{x} + 1} \cdot \frac{1}{2\sqrt{x}} \)
\( = \frac{1}{2x + 2\sqrt{x}} \)
Now substitute `x = 4`
\( = \frac{1}{2(4) + 2\sqrt{4}} \)
\( = \frac{1}{12} \)
Hence, slope of the tangent on the curve \( f(x) = \ln(\sqrt{x} + 1) \) at `x = 4` is `1/12`.
Example `4`. Calculate the derivative of the function \( f(x) = \ln(3x) \).
Solution:
Given function: \( f(x) = \ln(3x) \)
To find the derivative \( f'(x) \), we'll use the chain rule:
\( f'(x) = \frac{d}{dx} \ln(u) \)
\( = \frac{1}{u} \cdot \frac{du}{dx} \)
where \( u = 3x \).
Now, we'll find \( \frac{du}{dx} \):
\( \frac{du}{dx} = \frac{d}{dx}(3x) \)
\( = 3 \)
Substitute \( u = 3x \) and \( \frac{du}{dx} = 3\),
\( f'(x) = \frac{1}{3x} \cdot 3 \)
\( = \frac{1}{x} \)
Example `5`. Calculate the derivative of the function \( g(x) = \ln(2x^2 + 1) \).
Solution:
Given function: \( g(x) = \ln(2x^2 + 1) \)
To find the derivative \( g'(x) \), again we'll use the chain rule:
\( g'(x) = \frac{d}{dx} \ln(u) \)
\( = \frac{1}{u} \cdot \frac{du}{dx} \)
where \( u = 2x^2 + 1 \).
Now, we'll find \( \frac{du}{dx} \):
\( \frac{du}{dx} = \frac{d}{dx}(2x^2 + 1) \)
\( = 4x \)
Substitute \( u = 2x^2 + 1 \) and \( \frac{du}{dx} = 4x\),
\( g'(x) = \frac{1}{2x^2 + 1} \cdot 4x \)
\( = \frac{4x}{2x^2 + 1} \)
Practice Problems
Q`1`. Calculate the derivative of the function \( f(x) = \ln(5x) \).
- \( \frac{1}{x} \)
- \( \frac{2}{x} \)
- \( \frac{5}{x} \)
- \( \frac{3}{x} \)
Answer: a
Q`2`. Calculate the derivative of the function \( g(x) = \ln(4x^3) \).
- \( \frac{4}{x} \)
- \( \frac{3}{x} \)
- \( \frac{12}{x} \)
- \( \frac{7}{x} \)
Answer: b
Q`3`. Calculate the derivative of the function \( f(x) = \ln(2x + 3) \).
- \( \frac{3}{2x + 3} \)
- \( \frac{2}{2x + 2} \)
- \( \frac{2}{2x + 3} \)
- \( \frac{5}{2x + 3} \)
Answer: c
Q`4`. Calculate the derivative of the function \( g(x) = \ln(4x^2 - 5x + 1) \).
- \( \frac{1}{4x^3 - 5x + 1} \cdot (8x - 5) \)
- \( \frac{1}{6x^2 - 5x + 1} \cdot (2x - 5) \)
- \( \frac{1}{5x^2 - 4x + 1} \cdot (8x - 1) \)
- \( \frac{1}{4x^2 - 5x + 1} \cdot (8x - 5) \)
Answer: d
Q`5`. Calculate the derivative of the function \( g(x) = \ln(1 - e^{-x}) \).
- \( \frac{1}{e^x - 1} \)
- \( \frac{2}{e^x - 1} \)
- \( \frac{1}{1 - e^{-2x}} \)
- \( \frac{6}{1 - e^{-x}} \)
Answer: a
Frequently Asked Questions
`1`. What is the derivative of `ln(x)`?
Answer: The derivative of the natural logarithm function \( \ln(x) \) with respect to \( x \) is \( \frac{d}{dx}\ln(x) = \frac{1}{x} \).
`2`. How do you find the derivative of `ln(u)` using the chain rule?
Answer: To find the derivative of \( \ln(u) \), where \( u \) is a function of \( x \), you use the chain rule. The derivative is given by \( \frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx} \), where \( \frac{du}{dx} \) represents the derivative of \( u \) with respect to \( x \).
`3`. What is the derivative of `ln(ax + b)`, where `a` and `b` are constants?
Answer: The derivative of \( \ln(ax + b) \) with respect to \( x \) is \( \frac{a}{ax + b} \). You can find this derivative using the chain rule.
`4`. How do you find critical points of a function involving ln(x)?
Answer: To find critical points, you first find the derivative of the function using the chain rule if necessary. Then, set the derivative equal to zero and solve for \( x \). These solutions represent potential critical points.
`5`. Can you explain how to use the derivative of `ln(x)` in optimization problems?
Answer: In optimization problems, you often seek to maximize or minimize a function. You can use the derivative of \( \ln(x) \) to find critical points, and then use the first or second derivative tests to determine if these points correspond to maximum or minimum values of the function.