Question

Using the Quotient Rule In Exercises \(7-12,\) use the Quotient Rule to find the derivative of the function. $$ h(x)=\frac{\sqrt{x}}{x^{3}+1} $$

Short answer

The derivative of \(h(x) = \frac{\sqrt{x}}{x^{3} + 1}\) is \(h'(x) = \frac{x^{5/2} - \frac{1}{2\sqrt{x}}}{(x^{3} + 1)^2}\)

Step-by-step solution

Identify the functions

Identify the function in the numerator and the function in the denominator of the given function. Here, \(f(x) = \sqrt{x} = x^{1/2}\) is the function in the numerator and \(g(x) = x^{3} + 1\) is the function in the denominator.

Differentiate the functions

Differentiate both functions. Use the power rule in each case: The derivative of \(f(x)\), denoted as \(f'(x)\), is \(\frac{1}{2}x^{-1/2}\) = \(\frac{1}{2\sqrt{x}}\), The derivative of \(g(x)\), denoted as \(g'(x)\), is \(3x^{2}\).

Apply the Quotient Rule

While applying the Quotient Rule to get the derivative of the main function out of the derivatives of its numerator and denominator:The Quotient Rule states that the derivative of a ratio of two functions is \(h'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))²}\),So, applying the Quotient Rule with the derivatives obtained above: \[h'(x) = \frac{(\frac{\sqrt{x}}{2}) \cdot (3x^{2}) - (\frac{1}{2\sqrt{x}}) \cdot (x^{3} + 1)}{(x^{3} + 1)^2}\] .

Simplify the equation

The equation can be further simplified to remove radicals from the denominator and organize: \[h'(x) = \frac{\frac{3x^{5/2}}{2} - (\frac{x^{3}}{2\sqrt{x}} + \frac{1}{2\sqrt{x}})}{(x^{3} + 1)^2}\] which is \[h'(x) = \frac{\frac{3x^{5/2}}{2} - \frac{x^{5/2}}{2} - \frac{1}{2\sqrt{x}}}{(x^{3} + 1)^2}\] Simplifying the equation further, \[h'(x) = \frac{x^{5/2} - \frac{1}{2\sqrt{x}}}{(x^{3} + 1)^2}\]

Key concepts

Understanding the Derivative

When dealing with functions, the concept of a "derivative" is integral. The derivative tells us how a function behaves, how its output values change as inputs change. Imagine you're driving; the speedometer measures how fast you’re going – this is much like a derivative measuring the rate of change. In this exercise, you want to find how the function \( h(x) = \frac{\sqrt{x}}{x^{3}+1} \) changes as \( x \) changes. To achieve this, we employ the "Quotient Rule," which is used when you have two functions in a division form. The derivative will reveal how the ratio of the numerator to the denominator changes with \( x \).

Applying the Power Rule

The "Power Rule" is a fundamental technique in differentiation that simplifies finding the derivative of functions. It states that the derivative of \( x^n \) is \( nx^{n-1} \). This rule simplifies complex calculations.For the given function, you have:

• Numerator: \( f(x) = \sqrt{x} = x^{1/2} \)
• Applying the Power Rule: \( f'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \)
• Denominator: \( g(x) = x^3 + 1 \)
• Applying the Power Rule: \( g'(x) = 3x^{2} \)

These derivatives help in applying the Quotient Rule effectively.

Role of the Numerator in Quotient Rule

In any division of functions, the "numerator" is essential. It's the function placed over the denominator. In this exercise, the numerator \( f(x) = \sqrt{x} \) is used to find the effective rate of change across this division. When applying the Quotient Rule, the derivative of the numerator \( f'(x) \) comes into play.Here's how it looks in the Quotient Rule calculation:

• The term involving numerator is \( g(x)f'(x) \)
• It's combined with the derivative of the denominator to calculate the overall derivative \( h'(x) \)

By understanding this, you can see how the numerator influences changes across the function.

Impact of the Denominator in Differentiation

Like the numerator, the "denominator" critically affects the calculation. It forms the base function over which the numerator operates. In this exercise, the denominator is \( g(x) = x^3 + 1 \). Differentiating this gives \( g'(x) = 3x^{2} \).When applying the Quotient Rule, the denominator (and its derivative) affects both the numerator's derivative and keeps the structure of the division intact:

• Denominator appears in the formula as \( (g(x))^2 \), maintaining balance in the differentiation.
• The derivative of the denominator \( g'(x) \) subtracted from the derivative of the numerator maintains accuracy in calculating \( h'(x) \).

Without the denominator's role, differentiation in quotients would lack precision.