Derivative of Cot x

We now understand the derivative of cot x. It is commonly represented as d/dx (cot x) or (cot x)', and its value is -csc²x. The function cot x has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below: Cotangent Function: (cot(x) = cos(x)/sin(x)). Quotient Rule: Rule for differentiating cot(x) (since it consists of cos(x)/sin(x)). Cosecant Function: csc(x) = 1/sin(x).

The derivative of cot x can be denoted as d/dx (cot x) or (cot x)'. The formula we use to differentiate cot x is: d/dx (cot x) = -csc²x (or) (cot x)' = -csc²x The formula applies to all x where sin(x) ≠ 0

We can derive the derivative of cot 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: By First Principle Using Chain Rule Using Product Rule We will now demonstrate that the differentiation of cot x results in -csc²x using the above-mentioned methods: By First Principle The derivative of cot x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of cot x using the first principle, we will consider f(x) = cot 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) = cot x, we write f(x + h) = cot (x + h). Substituting these into equation (1), f'(x) = limₕ→₀ [cot(x + h) - cot x] / h = limₕ→₀ [ [cos (x + h) / sin (x + h)] - [cos x / sin x] ] / h = limₕ→₀ [ [cos (x + h) sin x - sin (x + h) cos x] / [sin x · sin(x + h)] ]/ h We now use the formula cos A sin B - sin A cos B = -sin (A - B). f'(x) = limₕ→₀ [ -sin (x + h - x) ] / [ h sin x · sin(x + h)] = limₕ→₀ [ -sin h ] / [ h sin x · sin(x + h)] = limₕ→₀ (-sin h)/ h · limₕ→₀ 1 / [sin x · sin(x + h)] Using limit formulas, limₕ→₀ (sin h)/ h = 1. f'(x) = -1 [ 1 / (sin x · sin(x + 0))] = -1/sin²x As the reciprocal of sine is cosecant, we have, f'(x) = -csc²x. Hence, proved. Using Chain Rule To prove the differentiation of cot x using the chain rule, We use the formula: Cot x = cos x/ sin x Consider f(x) = cos x and g (x)= sin x So we get, cot x = f(x)/ g(x) By quotient rule: d/dx [f(x) / g(x)] = [f '(x) g(x) - f(x) g'(x)] / [g(x)]² … (1) Let’s substitute f(x) = cos x and g (x) = sin x in equation (1), d/ dx (cot x) = [(-sin x) (sin x)- (cos x) (cos x)]/ (sin x)² (-sin²x - cos²x)/ sin²x …(2) Here, we use the formula: (sin²x) + (cos²x) = 1 (Pythagorean identity) Substituting this into (2), d/dx (cot x) = -1/ (sin x)² Since csc x = 1/sin x, we write: d/dx(cot x) = -csc²x Using Product Rule We will now prove the derivative of cot x using the product rule. The step-by-step process is demonstrated below: Here, we use the formula, Cot x = cos x/ sin x cot x = (cos x). (sin x)⁻¹ Given that, u = cos x and v = (sin x)⁻¹ Using the product rule formula: d/dx [u.v] = u'. v + u. v' u' = d/dx (cos x) = -sin x. (substitute u = cos x) Here we use the chain rule: v = (sin x)⁻¹ = (sin x)⁻¹ (substitute v = (sin x)⁻¹) v' = -1. (sin)⁻². d/dx (sin x) v' = -cos x/ (sin x)² Again, use the product rule formula: d/dx (cot x) = u'. v + u. v' Let’s substitute u = cos x, u' = -sin x, v = (sin x)⁻¹, and v' = -cos x/ (sin x)² When we simplify each term: We get, d/dx (cot x) = -1 - cos²x / (sin x)² cos²x/ (sin x)² = cot²x (we use the identity sin²x + cos²x =1) Thus: d/dx (cot x) = -1 - cot²x Since, -1 - cot²x = -csc²x d/dx (cot x) = -csc²x.

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 cot (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 cot(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 the x is 0, the derivative of cot x = -csc²(0), which is undefined because cot(x) has a vertical asymptote there. When the x is π/4, the derivative of cot x = -csc²(π/4), which is -2.

Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x. Cotangent Function: The cotangent function is one of the primary six trigonometric functions and is written as cot x. Cosecant Function: A trigonometric function that is the reciprocal of the sine function. It is typically represented as csc x. First Derivative: It is the initial result of a function, which gives us the rate of change of a specific function. Asymptote: The function goes near a line without intersecting or crossing it. This line is known as an asymptote.