Problem 29 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$y=\frac{1+\ln t}{1-\ln t}$$

Step-by-Step Solution

Answer

\( \frac{dy}{dt} = \frac{2}{t(1 - \ln t)^2} \)

Step 1: Identify the differentiation rule

Given the function \( y = \frac{1 + \ln t}{1 - \ln t} \), we need to find its derivative with respect to \( t \). This is a quotient of two functions, so we'll use the quotient rule for differentiation. The quotient rule states that for \( y = \frac{u}{v} \), the derivative \( \frac{dy}{dt} = \frac{v \cdot \frac{du}{dt} - u \cdot \frac{dv}{dt}}{v^2} \), where \( u = 1 + \ln t \) and \( v = 1 - \ln t \).

Step 2: Differentiate the numerator and denominator

Calculate the derivatives of the numerator \( u = 1 + \ln t \) and denominator \( v = 1 - \ln t \). The derivative of \( u \) with respect to \( t \) is \( \frac{du}{dt} = \frac{d}{dt} (1 + \ln t) = \frac{1}{t} \). Similarly, the derivative of \( v \) with respect to \( t \) is \( \frac{dv}{dt} = \frac{d}{dt} (1 - \ln t) = -\frac{1}{t} \).

Step 3: Apply the quotient rule

Substitute \( \frac{du}{dt} \) and \( \frac{dv}{dt} \) into the quotient rule: \[ \frac{dy}{dt} = \frac{(1 - \ln t) \cdot \left(\frac{1}{t}\right) - (1 + \ln t) \cdot \left(-\frac{1}{t}\right)}{(1 - \ln t)^2} \].

Step 4: Simplify the expression

Simplify the expression in the numerator: \( (1 - \ln t) \left(\frac{1}{t}\right) + (1 + \ln t) \left(\frac{1}{t}\right) = \frac{1 - \ln t + 1 + \ln t}{t} = \frac{2}{t} \).

Step 5: Final expression for the derivative

The lead fraction becomes \( \frac{dy}{dt} = \frac{\frac{2}{t}}{(1 - \ln t)^2} \). Simplifying this, we can write the final derivative as \( \frac{dy}{dt} = \frac{2}{t(1 - \ln t)^2} \).

Key Concepts

Quotient Rule

When you have a function that is the division of two other functions, the quotient rule helps you find the derivative. It's perfect for scenarios like ours, where you have a fraction to deal with. Think of the quotient rule as your guide in breaking down derivatives of these fractions without stressing over each element. The rule itself is simple: if you have a function \( y = \frac{u}{v} \), where both \( u \) and \( v \) are functions of a variable, the derivative \( \frac{dy}{dt} \) is expressed as \[ \frac{dy}{dt} = \frac{v \cdot \frac{du}{dt} - u \cdot \frac{dv}{dt}}{v^2} \]. This expression tells you to:

• Focus on finding derivatives of the top \((u)\) and bottom \((v)\) functions separately.
• Multiply the derivative of \( u \) by \( v \), and the derivative of \( v \) by \( u \) - then subtract those results.
• Divide the outcome by \( v^2 \), which is just the square of the bottom function.

Once you get comfortable with these steps, solving any derivative using the quotient rule becomes a straightforward task. In our original example, this rule helped us derive \( \frac{dy}{dt} = \frac{2}{t(1 - \ln t)^2} \).
Even though it looks complex at first, step by step, it becomes clear and logical.

Logarithmic Differentiation

Logarithmic differentiation is a powerful technique in calculus, particularly useful when dealing with functions that include logarithms. The essence of this method is to take advantage of logarithmic properties to simplify differentiation. When a logarithm is part of a function, it often transforms multiplication into addition or division into subtraction, turning what looks like a difficult differentiation problem into a simpler one. For example, suppose you have a function \( f(t) = \ln t \). The derivative \( \frac{d}{dt}(\ln t) \) simplifies to \( \frac{1}{t} \). This property holds true under most sub-conditions, making logarithmic differentiation a handy tool to recognize and use in calculus. There is no need to fear complex logarithmic elements – they simply need careful handling. Breaking them down with this technique changes the way you might perceive these types of problems. Knowing when to apply logarithmic differentiation can indeed make life much easier when studying or working in the realm of calculus.

Calculus Differentiation Techniques

Calculus offers a variety of differentiation techniques that help solve tough problems. Whether you're handling polynomial, trigonometric, exponential, or logarithmic functions, each differentiation technique makes complex issues manageable. A few of the major techniques include:

• The power rule, useful for polynomials, where you multiply the power by the coefficient and then decrease the power by one.
• The product rule, helping with multiplication of two functions and differentiated by \((uv)' = u'v + uv'\).
• The quotient rule, specifically for fractions and already discussed in detail above.

These rules are fundamental.More specialized methods like implicit differentiation and logarithmic differentiation help when things get complex – say, when a variable is buried inside a logarithm or an exponential function. These techniques give you tools to simplify and solve problems that might initially seem tricky. Making the right choice on which technique to apply comes with practice and understanding of different scenarios. As you study, honing your skill in applying these techniques will give you confidence in mastering advanced calculus.