Question

Differentiate the following functions: \(\frac{3-4 x+x^{3}}{x^{2}}\)

Short answer

Question: Find the derivative of the function \(f(x) = \frac{3-4x+x^3}{x^2}\) with respect to \(x\). Answer: The derivative of the function \(f(x) = \frac{3-4x+x^3}{x^2}\) with respect to \(x\) is \(f'(x) = \frac{x^4 + 4x^2 - 6x}{x^4}\).

Step-by-step solution

Identify g(x) and h(x)

In the given function \(\frac{3-4x+x^3}{x^2}\), let \(g(x) = 3-4x+x^3\) and \(h(x) = x^2\). Our goal is to find the derivative of this function using the Quotient Rule.

Differentiate g(x)

Differentiate \(g(x) = 3-4x+x^3\) with respect to \(x\) to find \(g'(x)\). Using the power rule for differentiation, we have: \(g'(x) = \frac{d}{dx}(3) - 4\frac{d}{dx}(x) + \frac{d}{dx}(x^3)\) \(g'(x) = 0 - 4\cdot 1 + 3x^2\) So, \(g'(x) = 3x^2 - 4\).

Differentiate h(x)

Now, differentiate \(h(x)=x^2\) with respect to \(x\) to find \(h'(x)\). Using the power rule, we have: \(h'(x) = \frac{d}{dx}(x^2)\) \(h'(x) = 2x\)

Apply the Quotient Rule to find the derivative of the function f(x)

Now that we have the derivatives of \(g(x)\) and \(h(x)\), we can apply the Quotient Rule: \(f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2}\) Now let's substitute the known values into the formula: \(f'(x) = \frac{(3x^2 - 4)(x^2) - (3 - 4x + x^3)(2x)}{(x^2)^2}\) Simplify the expression: \(f'(x) = \frac{3x^4 - 4x^2 - 2x(3 - 4x + x^3)}{x^4}\) \(= \frac{3x^4 - 4x^2 - 6x + 8x^2 - 2x^4}{x^4}\) Combine like terms: \(f'(x) = \frac{x^4 + 4x^2 - 6x}{x^4}\) Now we have the derivative of the given function with respect to \(x\).

Key concepts

Quotient Rule

The Quotient Rule is a nifty tool in calculus for taking the derivative of a function that is the quotient of two other functions. When you have a function set up as \ \( \frac{u(x)}{v(x)} \ \), where both \( u(x) \) and \( v(x) \) are themselves functions of \( x \), you'll employ the Quotient Rule to find the derivative. The formula is:

• Formula: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \)

In words, this means you:

• Take the derivative of the numerator, \( u(x) \), and multiply by the original denominator, \( v(x) \).
• Subtract the product of the original numerator, \( u(x) \), and the derivative of the denominator, \( v(x) \).
• Place this entire expression over the square of the original denominator, \( v(x)^2 \).

This systematic process helps to accurately find the derivative of quotients in functions, offering precise and usable results, crucial for delving deeper into the world of calculus.

Power Rule

The Power Rule is a fundamental component of differentiation. It's a straightforward rule used to find the derivative of a function that is a power of \( x \). The rule states that if you have a function \( f(x) = x^n \), then its derivative \( f'(x) \) is:

• Formula: \( \frac{d}{dx}(x^n) = nx^{n-1} \)

This means that you multiply the entire function by the exponent \( n \) and reduce the power by one. For example, the derivative of \( x^3 \) becomes \( 3x^2 \), and \( x^2 \) becomes \( 2x \).

• This rule is particularly useful for polynomial functions.
• It greatly simplifies the process of differentiation, turning potentially complex calculations into easy steps.

Using the Power Rule is intuitive when applied correctly, transforming any power function into its derivative form in a snap.

Derivatives

Derivatives are a primary concept in calculus used to study how a function changes as its input changes. Essentially, the derivative represents the rate of change or the slope of the function at any given point. Finding derivatives answers questions about velocity, growth rates, and anything involving change.
Here's how derivatives work and why they're so vital:

• Definition: The derivative of a function \( f(x) \) is often denoted as \( f'(x) \) or \( \frac{df}{dx} \).
• Purpose: It aims to capture how a slight change in \( x \) produces a change in \( f(x) \).

Derivatives are calculated through rules like the Quotient and Power Rules, connecting the dots in calculus efficiently. They're not limited to simple functions; derivatives work for trigonometric, exponential, and logarithmic functions as well. Mastery of derivatives allows you to unlock the full potential of mathematical analysis and problem-solving.