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Last updated on September 26, 2025
We use the derivative of ln(sin x), which is cot(x), as a tool for understanding how the logarithmic function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now discuss the derivative of ln(sin x) in detail.
We now understand the derivative of ln(sin x). It is commonly represented as d/dx (ln(sin x)) or (ln(sin x))', and its value is cot x. The function ln(sin x) has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Logarithmic Function: ln(x) is the natural logarithm function.
Chain Rule: Rule for differentiating ln(sin x) (since it involves a composition of functions).
Cotangent Function: cot(x) = 1/tan(x).
The derivative of ln(sin x) can be denoted as d/dx (ln(sin x)) or (ln(sin x))'. The formula we use to differentiate ln(sin x) is: d/dx (ln(sin x)) = cot x (or) (ln(sin x))' = cot x
The formula applies to all x where sin(x) > 0
We can derive the derivative of ln(sin x) using proofs. To show this, we will use the trigonometric identities along with the rules of differentiation. There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of ln(sin x) results in cot x using the above-mentioned methods:
The derivative of ln(sin x) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of ln(sin x) using the first principle, we will consider f(x) = ln(sin x). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = ln(sin x), we write f(x + h) = ln(sin(x + h)). Substituting these into equation (1), f'(x) = limₕ→₀ [ln(sin(x + h)) - ln(sin x)] / h = limₕ→₀ ln([sin(x + h) / sin x]) / h Using the logarithmic identity ln(a) - ln(b) = ln(a/b), f'(x) = limₕ→₀ [ln(1 + (sin(x + h) - sin x)/sin x)] / h Using the limit definition, f'(x) = limₕ→₀ [(sin(x + h) - sin x)/(h sin x)] Using the identity sin(A + h) - sin A = 2 cos((2A + h)/2) sin(h/2), f'(x) = limₕ→₀ [cos(x + h/2) sin(h/2)/sin x] / (h/2) Using limit formulas, limₕ→₀ sin(h/2)/(h/2) = 1. f'(x) = cos x/sin x = cot x.
To prove the differentiation of ln(sin x) using the chain rule, We use the formula: Let u = sin x, then ln(u) Differentiate using the chain rule: d(ln(u))/dx = 1/u * du/dx So, d(ln(sin x))/dx = 1/sin x * cos x = cot x.
We will now prove the derivative of ln(sin x) using the product rule. Let y = ln(sin x) = ln u, where u = sin x. Differentiate using the chain rule: dy/dx = 1/u * du/dx = 1/sin x * cos x = cot x.
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky.
To understand them better, think of a car where the speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like ln(sin x).
For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′ (x) Similarly, the third derivative, f′′′(x) is the result of the second derivative and this pattern continues.
For the nth Derivative of ln(sin x), we generally use fⁿ(x) for the nth derivative of a function f(x) which tells us the change in the rate of change. (continuing for higher-order derivatives).
When x is 0, the derivative is undefined because ln(sin x) is undefined for sin x = 0.
When x is π/2, the derivative of ln(sin x) = cot(π/2), which is 0.
Students frequently make mistakes when differentiating ln(sin x). These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Calculate the derivative of ln(sin x)·cos x
Here, we have f(x) = ln(sin x)·cos x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = ln(sin x) and v = cos x. Let’s differentiate each term, u′ = d/dx (ln(sin x)) = cot x v′ = d/dx (cos x) = -sin x Substituting into the given equation, f'(x) = (cot x)·(cos x) + (ln(sin x))·(-sin x) Let’s simplify terms to get the final answer, f'(x) = cot x cos x - ln(sin x) sin x Thus, the derivative of the specified function is cot x cos x - ln(sin x) sin x.
We find the derivative of the given function by dividing the function into two parts.
The first step is finding its derivative and then combining them using the product rule to get the final result.
A certain wave pattern in a physics experiment is represented by the function y = ln(sin x) where y represents the intensity of the wave at a point x. If x = π/3, measure the rate of change of intensity.
We have y = ln(sin x) (wave intensity)...(1) Now, we will differentiate the equation (1) Take the derivative ln(sin x): dy/dx = cot x Given x = π/3 (substitute this into the derivative) cot(π/3) = 1/tan(π/3) cot(π/3) = 1/(√3) Hence, we get the rate of change of intensity at x = π/3 as 1/√3.
We find the rate of change of intensity at x = π/3 as 1/√3, which means that at a given point, the intensity of the wave would change at this rate.
Derive the second derivative of the function y = ln(sin x).
The first step is to find the first derivative, dy/dx = cot x...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [cot x] Here we use the derivative of cot x, d²y/dx² = -csc²x Therefore, the second derivative of the function y = ln(sin x) is -csc²x.
We use the step-by-step process, where we start with the first derivative.
We then differentiate cot x to find the second derivative.
Prove: d/dx (ln(sin²x)) = 2 cot x.
Let’s start using the chain rule: Consider y = ln(sin²x) = 2 ln(sin x) To differentiate, we use the chain rule: dy/dx = 2 d/dx (ln(sin x)) Since the derivative of ln(sin x) is cot x, dy/dx = 2 cot x Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation.
Then, we replace the derivative of ln(sin x) with cot x.
As a final step, we simplify to derive the equation.
Solve: d/dx (ln(sin x)/x)
To differentiate the function, we use the quotient rule: d/dx (ln(sin x)/x) = (d/dx (ln(sin x))·x - ln(sin x)·d/dx(x))/x² We will substitute d/dx (ln(sin x)) = cot x and d/dx (x) = 1 = (cot x·x - ln(sin x)·1)/x² = (x cot x - ln(sin x))/x² Therefore, d/dx (ln(sin x)/x) = (x cot x - ln(sin x))/x²
In this process, we differentiate the given function using the quotient rule.
As a final step, we simplify the equation to obtain the final result.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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