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Chapter 2: Problem 10

Use the Quotient Rule to differentiate the function. \(h(s)=\frac{s}{\sqrt{s}-1}\)

Short Answer

Expert verified
The derivative of the function \(h(s)=\frac{s}{\sqrt{s}-1}\) is given by \(h'(s)=\frac{1}{2}-\frac{1}{2\sqrt{s}}\).

Step by step solution

01

Identify the numerator and the denominator

The numerator of this function is \(s\) and the denominator is \(\sqrt{s}-1\).
02

Derive the numerator and the denominator

The derivative of \(s\) with respect to \(s\) is \(1\) and the derivative of \(\sqrt{s}-1\) with respect to \(s\) is \(\frac{1}{2\sqrt{s}}\).
03

Apply the Quotient Rule

By the Quotient Rule, the derivative \(h'(s)\) is given by \[h'(s)=\frac{\sqrt{s}-1}{2s\sqrt{s}}\]
04

Simplify the result

The derivative \(h'(s)\) simplifies to \[h'(s)=\frac{1}{2}-\frac{1}{2\sqrt{s}}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiate Rational Functions
Rational functions are a key topic in calculus, where they are recognized as ratios of polynomials. When differentiating rational functions like the example of
\(h(s) = \frac{s}{\sqrt{s}-1}\)
, the Quotient Rule is a critical technique that comes into play. To apply this rule effectively, one must first understand its formula: \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \), where \(u\) and \(v\) are functions of \(x\), and \(u'\) and \(v'\) are their respective derivatives.
Derivative of Square Root Functions
When dealing with square root functions, differentiation may seem daunting. However, the process adheres to the same principles that govern all differentiable functions, with the addition of paying attention to the chain rule. The derivative of a square root, which is a power function \(\sqrt{x} = x^{1/2}\), can be found using the power rule.
Applying this to the denominator of our example function \( \sqrt{s} - 1 \), and using the power rule \( \frac{d}{dx}(x^{n}) = nx^{n-1} \), we find the derivative to be \( \frac{1}{2\sqrt{s}} \). This is a crucial step to correctly applying the Quotient Rule in the given problem.
Calculus
Calculus is the branch of mathematics that deals with rates of change and accumulation. Differentiation, which we're focused on here, is a core concept in calculus. It allows us to compute the rate at which a function changes at any point. For a deep understanding, one should study the limits, the foundation of derivatives, and the various rules of differentiation like the Quotient Rule, Product Rule, and Chain Rule. These techniques are vital when working with complex functions, helping students to handle problems involving rates of change more intuitively.
Derivative Simplification
After applying differentiation rules, the resulting expressions can often be complex and require simplification. Such simplification helps in better understanding the behavior of the derivative function. In the context of our exercise, once the Quotient Rule is applied, simplifying the expression \(h'(s) = \frac{\sqrt{s}-1}{2s\sqrt{s}}\) to \(h'(s) = \frac{1}{2} - \frac{1}{2\sqrt{s}}\)
makes it more succinct and easier to analyze. This is an exemplary instance of derivative simplification which involves arithmetic skills and algebraic manipulation to reduce the derivative to its simplest form.

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Most popular questions from this chapter

Let \(u\) be a differentiable function of \(x\). Use the fact that \(|u|=\sqrt{u^{2}}\) to prove that $$ \frac{d}{d x}[|u|]=u^{\prime} \frac{u}{|u|}, \quad u \neq 0. $$

Find the points at which the graph of the equation has a vertical or horizontal tangent line. $$ 4 x^{2}+y^{2}-8 x+4 y+4=0 $$

Verify that the two families of curves are orthogonal where \(C\) and \(K\) are real numbers. Use a graphing utility to graph the two families for two values of \(C\) and two values of \(K\). $$ x y=C, \quad x^{2}-y^{2}=K $$

Find \(d y / d x\) by implicit differentiation and evaluate the derivative at the given point. $$ x \cos y=1, \quad\left(2, \frac{\pi}{3}\right) $$

Tangent Lines Find equations of both tangent lines to the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\) that passes through the point \((4,0)\).
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