Question

Find the derivative of each function.. $$v=\sqrt{\frac{1+2 t}{1-2 t}}$$

Short answer

\(v' = \frac{2}{\sqrt{\frac{1+2t}{1-2t}}(1-2t)^2}\)

Step-by-step solution

Simplify the function if possible

In this case, the function is already in its simplest form, so we do not need to do any simplification before differentiating.

Identify the outer function and inner function (Chain Rule)

To differentiate the given function, we will use the Chain Rule. The outer function is the square root function, and the inner function is the fraction \(\frac{1 + 2t}{1 - 2t}\). So if we let \(u = \frac{1+2t}{1-2t}\), then \(v = \sqrt{u}\).

Differentiate the outer function w.r.t. the inner function

The derivative of the outer function \(v = \sqrt{u}\) with respect to \(u\) is \(\frac{1}{2}\sqrt{u}^{-1} = \frac{1}{2}\sqrt{u}^{-1} = \frac{1}{2\sqrt{u}}\).

Differentiate the inner function w.r.t. \(t\)

The derivative of \(u = \frac{1+2t}{1-2t}\) with respect to \(t\) is obtained using the quotient rule: if \(u = \frac{f}{g}\), then \(u' = \frac{f'g - fg'}{g^2}\). By applying this rule, we get \(u' = \frac{(2)(1-2t) - (1+2t)(-2)}{(1-2t)^2}\) which simplifies to \(\frac{4}{(1-2t)^2}\).

Apply the Chain Rule

Now, combine the results of Step 3 and Step 4 to calculate \(v'\). The derivative \(v' = \frac{dv}{dt} = \frac{dv}{du} \cdot \frac{du}{dt}\), thus \(v' = \frac{1}{2\sqrt{u}} \cdot \frac{4}{(1-2t)^2}\).

Replace \(u\) back with \(\frac{1+2t}{1-2t}\)

Finally, substitute \(u\) back into the derivative to get the derivative in terms of \(t\): \(v' = \frac{1}{2\sqrt{\frac{1+2t}{1-2t}}} \cdot \frac{4}{(1-2t)^2} = \frac{2}{\sqrt{\frac{1+2t}{1-2t}}(1-2t)^2}\). This is the final form of the derivative of the original function.

Key concepts

Chain Rule in Calculus

Understanding the chain rule is essential when dealing with functions that are composed of other functions. Simply put, the chain rule is used in calculus to differentiate composite functions.

A composite function can be thought of as a combination of two or more functions where the output of one function becomes the input of another. To differentiate a composite function, we follow the chain rule, which says that the derivative of a composite function is the derivative of the outer function multiplied by the derivative of the inner function.

Breaking Down the Process

Let's consider our function from the original exercise, \(v = \sqrt{\frac{1+2t}{1-2t}}\). This function is the composition of a square root function and a fractional function. First, we identify the inner function \(u = \frac{1+2t}{1-2t}\) and the outer function, which is the square root of \(u\).

By applying the chain rule, we differentiate the outer function with respect to the inner function, and then multiply it by the derivative of the inner function with respect to \(t\). This step-by-step approach breaks down the composite function into simpler parts, making differentiation more manageable.

• The chain rule formula can be stated as \(\frac{dv}{dt} = \frac{dv}{du} \cdot \frac{du}{dt}\).
• In our example, \(\frac{dv}{du} = \frac{1}{2\sqrt{u}}\) and \(\frac{du}{dt}\) involves applying the quotient rule, which we'll explore in the next section.

This method not only simplifies complex differentiation but also ensures a clear understanding of the function's components and their relationships.

Quotient Rule for Derivatives

The quotient rule is a technique used to differentiate functions that are expressed as the quotient of two other functions. If we have a function \(u\) that is the division of two functions, \(u = \frac{f(t)}{g(t)}\), the quotient rule tells us how to find the derivative \(u'\).

The quotient rule formula is expressed as \(u' = \frac{f'g - fg'}{g^2}\), where \(f'\) and \(g'\) are the derivatives of the functions \(f\) and \(g\) with respect to \(t\).

Implementing the Quotient Rule

In our exercise example, we defined \(u\) as \(u = \frac{1+2t}{1-2t}\). To find \(u'\), we need to determine the derivatives of the numerator and the denominator separately:

• The derivative of the numerator \(1+2t\) is simply 2, while
• The derivative of the denominator \(1-2t\) is -2.

Applying the quotient rule: \(u' = \frac{(2)(1-2t) - (1+2t)(-2)}{(1-2t)^2}\), which simplifies to \(\frac{4}{(1-2t)^2}\).

The quotient rule is particularly handy when dealing with rational functions like the one in our exercise. It provides a systematic way to handle the differentiation of fractions where both the numerator and denominator are functions of the variable.

Step-by-Step Differentiation

Approaching differentiation in a step-by-step manner is a great strategy for solving complex problems. By breaking down the process into smaller, more manageable steps, students can tackle calculus problems that may at first seem daunting.

Each step in the differentiation process applies different rules or properties of derivatives, allowing for the piece-by-piece solving of the original problem. Let's revisit the example to illustrate this method.

Completing the Differentiation Process

After applying the chain rule and the quotient rule to our function \(v\), we arrived at \(v' = \frac{1}{2\sqrt{u}} \cdot \frac{4}{(1-2t)^2}\). The final step is to substitute the inner function \(u\) back into our derivative, obtaining the derivative in terms of the original variable \(t\).

This substitution gives us the final differentiated expression: \[v' = \frac{2}{\sqrt{\frac{1+2t}{1-2t}}(1-2t)^2}\].

The step-by-step differentiation approach is not only systematic but it also provides a clear pathway for understanding each action taken during the process. By focusing on individual steps, students can verify and understand the rationale behind each part of the derivative, thus reinforcing their conceptual understanding of the differentiation process.