Derivative of 5/x

We now understand the derivative of 5/x.

It is commonly represented as d/dx (5/x) or (5/x)', and its value is -5/x².

The function 5/x has a clearly defined derivative, indicating it is differentiable within its domain.

The key concepts are mentioned below:

Function: 5/x is a basic rational function.

Power Rule: Rule for differentiating xⁿ.

Negative Exponent: Using x^-1 to represent 1/x.

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

The formula we use to differentiate 5/x is: d/dx (5/x) = -5/x² (or) (5/x)' = -5/x².

The formula applies to all x where x ≠ 0.

We can derive the derivative of 5/x using proofs.

To show this, we will use algebraic manipulation along with the rules of differentiation.

There are several methods we use to prove this, such as:

By First Principle

Using Power Rule

Using Quotient Rule

We will now demonstrate that the differentiation of 5/x results in -5/x² using the above-mentioned methods:

By First Principle

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

To find the derivative of 5/x using the first principle, we will consider f(x) = 5/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) = 5/x, we write f(x + h) = 5/(x + h).

Substituting these into equation (1), f'(x) = limₕ→₀ [5/(x + h) - 5/x] / h = limₕ→₀ [5x - 5(x + h)] / [h(x)(x + h)] = limₕ→₀ [-5h] / [h(x)(x + h)] = limₕ→₀ [-5] / [x(x + h)] = -5/x²

Hence, proved.

Using Power Rule

To prove the differentiation of 5/x using the power rule, We rewrite the function as: 5/x = 5x^-1.

Using the power rule, if y = xⁿ, then dy/dx = nx^(n-1).

So we get, d/dx (5x^-1) = 5(-1)x^(-1-1) = -5x^-2.

Therefore, d/dx (5/x) = -5/x².

Using Quotient Rule

We will now prove the derivative of 5/x using the quotient rule.

The step-by-step process is demonstrated below:

Here, we use the formula, 5/x = 5 * (1/x).

Given that u = 5 and v = x⁻¹,

Using the quotient rule formula: d/dx [u/v] = [u'v - uv']/v². u' = 0 (since u = 5 is a constant) v' = d/dx (x⁻¹) = -x⁻² (substitute v = x⁻¹).

Using the quotient rule, d/dx (5/x) = [0 * x⁻¹ - 5 * (-x⁻²)] / (x⁻¹)² = (5x⁻²)/x⁻² = -5/x².

Hence, d/dx (5/x) = -5/x².

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 5/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 5/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 5/x is undefined at this point. When x is positive or negative, the derivative of 5/x = -5/x², indicating a negative slope.

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

• Power Rule: A rule used to differentiate expressions of the form xⁿ, which results in nx^(n-1).

• Quotient Rule: A rule used to differentiate functions that are ratios of two other functions.

• Undefined: A term used to describe a point where a function does not exist or is not continuous.

• Negative Exponent: A mathematical notation where x^-n is equivalent to 1/xⁿ.