Derivative of 1/u

We now understand the derivative of 1/u. It is commonly represented as d/du (1/u) or (1/u)', and its value is -1/u². The function 1/u has a clearly defined derivative, indicating it is differentiable within its domain.

The key concepts are mentioned below:

Reciprocal Function: (1/u).

Power Rule: Rule for differentiating 1/u (since it can be written as u⁻¹).

Negative Power: The derivative of a negative power involves multiplying by the negative exponent.

The derivative of 1/u can be denoted as d/du (1/u) or (1/u)'.

The formula we use to differentiate 1/u is: d/du (1/u) = -1/u² (or) (1/u)' = -1/u²

The formula applies to all u where u ≠ 0.

We can derive the derivative of 1/u using proofs. To show this, we will use algebraic identities along with the rules of differentiation. There are several methods we use to prove this, such as:

1. By First Principle
2. Using Power Rule
3. Using Quotient Rule

We will now demonstrate that the differentiation of 1/u results in -1/u² using the above-mentioned methods:

By First Principle

The derivative of 1/u can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.

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

Given that f(u) = 1/u, we write f(u + h) = 1/(u + h).

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

As h approaches 0, f'(u) = -1/u²

Hence, proved.

Using Power Rule

To prove the differentiation of 1/u using the power rule, We rewrite the function: 1/u = u⁻¹

Using the power rule: d/du (uⁿ) = n·uⁿ⁻¹ d/du (u⁻¹) = -1·u⁻² Thus, d/du (1/u) = -1/u²

Using Quotient Rule

We will now prove the derivative of 1/u using the quotient rule. The step-by-step process is demonstrated below:

Here, we use the identity, 1/u = 1 / u

Given that, f(u) = 1 and g(u) = u

Using the quotient rule formula: d/du [f(u) / g(u)] = [f '(u) g(u) - f(u) g'(u)] / [g(u)]²… (1)

Let’s substitute f(u) = 1 and g(u) = u in equation (1), d/du (1/u) = [0·u - 1·1] / u² = -1/u²

Thus, d/du (1/u) = -1/u².

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 1/u.

For the first derivative of a function, we write f′(u), 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′′(u). Similarly, the third derivative, f′′′(u), is the result of the second derivative, and this pattern continues.

For the nth Derivative of 1/u, we generally use fⁿ(u) for the nth derivative of a function f(u), which tells us the change in the rate of change (continuing for higher-order derivatives).

When u = 0, the derivative is undefined because 1/u has a vertical asymptote there. When u = 1, the derivative of 1/u = -1/1², which is -1.

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

• Reciprocal Function: A function that involves taking the reciprocal of a variable, written as 1/u.

• Power Rule: A basic rule of differentiation used to find the derivative of functions of the form uⁿ.

• Chain Rule: A rule for finding the derivative of compositions of functions.

• Undefined Point: A point in the domain of a function where the function does not exist or is not continuous.