Derivative of 5tanx

We now understand the derivative of 5tanx. It is commonly represented as d/dx (5tanx) or (5tanx)', and its value is 5sec²x. The function 5tanx has a clearly defined derivative, indicating it is differentiable within its domain.

The key concepts are mentioned below:

Tangent Function: (tan(x) = sin(x)/cos(x)).

Quotient Rule: Rule for differentiating tan(x) (since it consists of sin(x)/cos(x)).

Secant Function: sec(x) = 1/cos(x).

The derivative of 5tanx can be denoted as d/dx (5tanx) or (5tanx)'.

The formula we use to differentiate 5tanx is: d/dx (5tanx) = 5sec²x (or) (5tanx)' = 5sec²x

The formula applies to all x where cos(x) ≠ 0.

We can derive the derivative of 5tanx 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 5tanx results in 5sec²x using the above-mentioned methods:

By First Principle

The derivative of 5tanx can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of 5tanx using the first principle, we will consider f(x) = 5tanx. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = 5tanx, we write f(x + h) = 5tan(x + h). Substituting these into equation (1), f'(x) = limₕ→₀ [5tan(x + h) - 5tanx] / h = 5 limₕ→₀ [tan(x + h) - tanx] / h = 5 limₕ→₀ [[sin(x + h) / cos(x + h)] - [sinx / cosx]] / h = 5 limₕ→₀ [[sin(x + h)cosx - cos(x + h)sinx] / [cosx · cos(x + h)]] / h We now use the formula sinAcosB - cosAsinB = sin(A-B). f'(x) = 5 limₕ→₀ [sinh] / [h cosx · cos(x + h)] = 5 limₕ→₀ [sinh] / [h cosx · cos(x + h)] = 5 limₕ→₀ (sinh)/h · limₕ→₀ 1 / [cosx · cos(x + h)] Using limit formulas, limₕ→₀ (sinh)/h = 1. f'(x) = 5 [1 / (cosx · cos(x + 0))] = 5/cos²x As the reciprocal of cosine is secant, we have, f'(x) = 5sec²x. Hence, proved.

Using Chain Rule

To prove the differentiation of 5tanx using the chain rule, We use the formula: Tanx = sinx/cosx Consider f(x) = sinx and g(x) = cosx So we get, tanx = 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) = sinx and g(x) = cosx in equation (1), d/dx (tanx) = [(cosx)(cosx) - (sinx)(-sinx)] / (cosx)² (cos²x + sin²x) / cos²x …(2) Here, we use the formula: (cos²x) + (sin²x) = 1 (Pythagorean identity) Substituting this into (2), d/dx (tanx) = 1 / (cosx)² Since secx = 1/cosx, we write: d/dx(tanx) = sec²x Thus, d/dx (5tanx) = 5sec²x

Using Product Rule

We will now prove the derivative of 5tanx using the product rule. The step-by-step process is demonstrated below: Here, we use the formula, 5tanx = 5(sinx/cosx) 5tanx = 5(sinx)(cosx)⁻¹ Given that, u = sinx and v = (cosx)⁻¹ Using the product rule formula: d/dx [u.v] = u'.v + u.v' u' = d/dx (sinx) = cosx. (substitute u = sinx) Here we use the chain rule: v = (cosx)⁻¹ = (cosx)⁻¹ (substitute v = (cosx)⁻¹) v' = -1(cos)⁻² d/dx (cosx) v' = sinx / (cosx)² Again, use the product rule formula: d/dx (5tanx) = 5[u'.v + u.v'] Let’s substitute u = sinx, u' = cosx, v = (cosx)⁻¹, and v' = sinx / (cosx)² When we simplify each term: We get, d/dx (5tanx) = 5(1 + sin²x / (cosx)²) sin²x / (cosx)² = tan²x (we use the identity sin²x + cos²x = 1) Thus: d/dx (5tanx) = 5(1 + tan²x) Since, 1 + tan²x = sec²x d/dx (5tanx) = 5sec²x.

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 5tan(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 5tan(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 π/2, the derivative is undefined because tan(x) has a vertical asymptote there.

When x is 0, the derivative of 5tanx = 5sec²(0), which is 5.

• Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x.

• Tangent Function: The tangent function is one of the primary six trigonometric functions and is written as tanx.

• Secant Function: A trigonometric function that is the reciprocal of the cosine function. It is typically represented as secx.

• First Derivative: It is the initial result of a function, which gives us the rate of change of a specific function.

• Quotient Rule: A rule used in calculus to find the derivative of the quotient of two functions.