Problem 14 - The Calculus with Analytic Geometry

Question

Find \(f^{(4)}(x)\) if \(f(x)=\frac{2}{x-1}\).

Step-by-Step Solution

Answer

f^{(4)}(x) = 120/(x-1)^7

Step 1 - Find the First Derivative

To find the first derivative of the function, use the quotient rule. For a function in the form ddx f(x)= super{2}×(x)x−1, where u = 2 and v = x - 1: f'(x) = uv' - vu'v^2 f'(x) = (2)'(x - 1) - (x - 1)'(2) / (x - 1)^2. Simplify to get: f'(x) = -2/(x - 1)^2

Step 2 - Find the Second Derivative

Apply the power rule and chain rule to get the second derivative. f''(x)=ddx−2x−1 ⟹super{2}×(−2)(x−1)−2(x−1)’/(x−1)4f''(x)=super{8}(x−1)3

Step 3 - Differentiate to Find the Third Derivative

f'''(x)=ddx8x–13which gives: f'''(x)=8×(−1)(x–1)−4(x–1)’/(x–1)6=−24/(x-1)5 .

Step 4 -Differentiate to Find the Fourth Derivative

This step involves differentiation again: f''''(x)=–24(x –1)( –5) which gives: f''''(x)=-24 (–5)(x-1)−6(x–1)■■=120/(x-1)7.

Key Concepts

quotient rule

The quotient rule is used to find the derivative of a ratio of two functions. If we have a function in the form \( \frac{u}{v} \), where both u and v are differentiable, the quotient rule states:
\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2}\] Here, u' and v' are the derivatives of u and v, respectively.

This rule is particularly useful when dealing with complex fractions in calculus. Let's see it in action using the given function \( f(x) = \frac{2}{x-1} \).

- Let \( u = 2 \) and \( v = x - 1 \).
- Then, the derivatives are \( u' = 0 \) and \( v' = 1 \).
Applying the quotient rule:
\[ f'(x) = \frac{ (x - 1) \cdot 0 - 2 \cdot 1 }{(x - 1)^2} = \frac{ -2 }{(x - 1)^2} \]
This gives us the first derivative.

power rule

The power rule is a simple but powerful tool for finding the derivative of functions of the form \[ x^n \] where n is any real number. The rule states:
\[ \frac{d}{dx} x^n = nx^{n-1} \]
This formula is foundational and can be applied repeatedly for higher-order derivatives.

In our exercise, we use the power rule when simplifying terms during differentiation. For example, when finding the second derivative:
\[ f''(x) = \frac{d}{dx} \left( \frac{-2}{(x-1)^2} \right) = -2 \frac{d}{dx} (x - 1)^{-2} \]
Using the power rule, this becomes:
\[ = -2 \cdot (-2) (x - 1)^{-3} = \frac{4}{(x-1)^3} \]
This technique greatly simplifies working with polynomial terms.

chain rule

The chain rule is essential for finding derivatives of composite functions. If a function y depends on u, and u depends on x, the chain rule allows us to find \frac{dy}{dx}\ as follows:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]
This approach is necessary when differentiating nested functions.

In the given exercise, the chain rule is applied to the second derivative and beyond. For example:
\[ f''(x) = \frac{d}{dx} \left( \frac{4}{(x-1)^3} \right)\, \text{we set} \ u=(x-1) \text{and} y = u^{-3} \]
\[ \frac{dy}{dx} = -3u^{-4} \cdot \frac{d}{dx} (x-1) = -3u^{-4} \cdot 1 = -3(x-1)^{-4}\]
When combined with other rules, the chain rule is vital for tackling complex derivatives.

calculus

Calculus is the branch of mathematics that deals with rates of change and accumulation. It has two main branches: differential calculus and integral calculus.

- Differential calculus focuses on rates of change and slopes of curves. The main tool here is the derivative, which measures how a function changes as its input changes.

- Integral calculus, on the other hand, deals with accumulation of quantities and the areas under and between curves. The integral is the primary concept here.

The exercise provided falls under differential calculus as it requires us to find higher-order derivatives of a given function.
Understanding and applying the right calculus rules—like the quotient rule, power rule, and chain rule—helps in efficiently solving these problems and deepening our grasp of how functions behave.