Derivative of 5cosx

We now understand the derivative of 5cosx. It is commonly represented as d/dx (5cosx) or (5cosx)', and its value is -5sinx.

The function 5cosx has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:

Cosine Function: (cos(x)).

Constant Rule: Rule for differentiating constants multiplied by a function.

Sine Function: sin(x) is the derivative of cos(x).

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

The formula we use to differentiate 5cosx is: d/dx (5cosx) = -5sinx (or) (5cosx)' = -5sinx

The formula applies to all x.

We can derive the derivative of 5cosx 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:

1. By First Principle
2. Using Chain Rule
3. Using Product Rule

We will now demonstrate that the differentiation of 5cosx results in -5sinx using the above-mentioned methods:

By First Principle

The derivative of 5cosx can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.

To find the derivative of 5cosx using the first principle, we will consider f(x) = 5cosx. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)

Given that f(x) = 5cosx, we write f(x + h) = 5cos(x + h).

Substituting these into equation (1), f'(x) = limₕ→₀ [5cos(x + h) - 5cosx] / h = limₕ→₀ 5[cos(x + h) - cosx] / h = 5limₕ→₀ [-2sin((x + x + h)/2)sin(h/2)] / h

Using the formula for the limit, sin(h/2)/(h/2) = 1 as h approaches 0, we have: f'(x) = 5[-2sin(x) * 1] = -5sinx. Hence, proved.

Using Chain Rule

To prove the differentiation of 5cosx using the chain rule, We use the formula:

5cosx = 5 * cosx Consider u(x) = 5 and v(x) = cosx.

Using the product rule: d/dx [u(x) * v(x)] = u'(x) * v(x) + u(x) * v'(x)

Let’s substitute u(x) = 5 and v(x) = cosx, d/dx (5cosx) = 0 * cosx + 5 * (-sinx) = -5sinx.

Using Product Rule

We will now prove the derivative of 5cosx using the product rule. The step-by-step process is demonstrated below: Here, we use the formula, 5cosx = (5).(cosx) Given that, u = 5 and v = cosx

Using the product rule formula: d/dx [u.v] = u'.v + u.v' u' = d/dx (5) = 0 v' = d/dx (cosx) = -sinx

Again, use the product rule formula: d/dx (5cosx) = u'v + uv' = 0 * cosx + 5 * (-sinx) = -5sinx.

Thus: d/dx (5cosx) = -5sinx.

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 5cos(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 5cos(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 π/2, the derivative is -5sin(π/2), which is -5. When the x is 0, the derivative of 5cosx = -5sin(0), which is 0.

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

• Cosine Function: The cosine function is one of the primary six trigonometric functions and is written as cosx.

• Sine Function: A trigonometric function that is the derivative of the cosine function. It is typically represented as sinx.

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

• Product Rule: A rule used to differentiate the product of two functions.