Problem 9 - Thomas Calculus

Question

Find the derivatives of the functions \(s=\frac{\sqrt{t}}{1+\sqrt{t}}\)

Step-by-Step Solution

Answer

The derivative is \( s' = \frac{1}{2\sqrt{t}(1+\sqrt{t})^2} \).

Step 1: Identify the Rule to Use

Since the function is a quotient of two expressions, we will use the Quotient Rule for differentiation. The Quotient Rule states that for a function \( s = \frac{u}{v} \), the derivative \( s' \) is given by \( s' = \frac{u'v - uv'}{v^2} \). In our case, \( u=\sqrt{t} \) and \( v=1+\sqrt{t} \).

Step 2: Differentiate the Numerator

The numerator of our function is \( u = \sqrt{t} = t^{1/2} \). To find \( u' \), use the power rule \( (t^n)' = nt^{n-1} \). Thus, \( u' = \frac{1}{2}t^{-1/2} = \frac{1}{2\sqrt{t}} \).

Step 3: Differentiate the Denominator

The denominator is \( v = 1 + \sqrt{t} = 1 + t^{1/2} \). Its derivative \( v' \) is 0 (derivative of constant 1) plus the derivative of \( t^{1/2} \), which is \( \frac{1}{2}t^{-1/2} = \frac{1}{2\sqrt{t}} \). Thus, \( v' = \frac{1}{2\sqrt{t}} \).

Step 4: Apply the Quotient Rule

Using the Quotient Rule: \( s' = \frac{u'v - uv'}{v^2} \).- Substitute \( u' = \frac{1}{2\sqrt{t}} \), \( u = \sqrt{t} \), \( v = 1 + \sqrt{t} \), and \( v' = \frac{1}{2\sqrt{t}} \) into the formula.- Compute: \( s' = \frac{\frac{1}{2\sqrt{t}}(1+\sqrt{t}) - \sqrt{t} \cdot \frac{1}{2\sqrt{t}}}{(1+\sqrt{t})^2} \).

Step 5: Simplify the Expression

Calculate the expression from Step 4:1. Simplify the terms: \( \frac{1}{2\sqrt{t}}(1+\sqrt{t}) = \frac{1}{2\sqrt{t}} + \frac{1}{2} \) and \( \sqrt{t} \cdot \frac{1}{2\sqrt{t}} = \frac{1}{2} \).2. Combine: \( s' = \frac{\left(\frac{1}{2\sqrt{t}} + \frac{1}{2} - \frac{1}{2}\right)}{(1+\sqrt{t})^2} = \frac{1}{2\sqrt{t}(1+\sqrt{t})^2} \).3. The final derivative is \( s' = \frac{1}{2\sqrt{t}(1+\sqrt{t})^2} \).

Key Concepts

Quotient Rule

The Quotient Rule is a technique used in calculus to find the derivative of a function that is the ratio of two differentiable functions. Suppose you have a function in the form of a fraction \( s = \frac{u}{v} \). In such cases, the derivative \( s' \) can be calculated using the formula:

• \( s' = \frac{u'v - uv'}{v^2} \)

This means you:

• First find the derivative of the numerator \( u' \).
• Next, find the derivative of the denominator \( v' \).
• Substitute these derivatives back into the formula.

It's essential to carefully carry out these calculations, as each part of the formula has specific roles. Always remember to square the denominator in the final expression, ensuring correct application of the rule. This technique is particularly useful when functions themselves are complex polynomials or contain square roots.

Power Rule

The Power Rule is one of the most fundamental rules in calculus for finding the derivative of terms containing powers of a variable. If you're presented with an expression like \( t^n \), you can apply the Power Rule, which states:

• \((t^n)' = nt^{n-1}\)

This means that you multiply the original exponent by the variable and decrease the exponent by one. For example, to differentiate \( \sqrt{t} \), note that it is equivalent to \( t^{1/2} \). Applying the Power Rule gives:

• \( (t^{1/2})' = \frac{1}{2}t^{-1/2} \)
• Rewriting this, you get \( \frac{1}{2\sqrt{t}} \)

This strategy simplifies the differentiation of polynomial functions and terms involving square roots, making it incredibly powerful when addressing such mathematical challenges.

Differentiation Techniques

Differentiation Techniques cover various methods used to find the derivative of functions. While the Power Rule and Quotient Rule are very popular, it's essential to apply them correctly based on the form of your function.

• For functions involving sums, you can differentiate each term separately, leveraging their individual rules.
• Practically, identifying the structure of the expression first is crucial to selecting the appropriate technique.

Consider the example \( s = \frac{\sqrt{t}}{1+\sqrt{t}} \). This function required the Quotient Rule due to its division form. Differentiating each component:

• For \( \sqrt{t} \), use the Power Rule.
• Constant terms simplify to zero.

This shows a practical combination of differentiation strategies, demonstrating their application beyond theoretical understanding. With practice, you’ll become adept at choosing and applying these methods effortlessly.