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) = csc² x (or) (-cot x)’ = csc² x

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/sin² x As the reciprocal of sine is cosecant, we have, f'(x) = csc² x.

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 (-sin² x + cos² x)/ sin² x …(2)

Here, we use the formula: (cos² x) - (sin² x) = cos(2x) (Trigonometric 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) = csc² x

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 -cos² x/ (sin x)² = -cot² x (we use the identity cos² x - sin² x =cos(2x)) Thus: d/dx (-cot x) = 1 + cot² x Since, 1 + cot² x = csc² x d/dx (-cot x) = csc² 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 -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 x is a multiple of π, the derivative is undefined because -cot(x) has a vertical asymptote there. When 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 for its negative counterpart.

• 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 approaches a line but never intersects or crosses it. This line is known as an asymptote.