Problem 50 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. $$ h(t)=\frac{\ln t}{1+t^{2}} $$

Step-by-Step Solution

Answer

The derivative of \( h(t) = \frac{\ln t}{1+t^2} \) is \( h'(t) = \frac{1 + t^2 - 2t^2 \ln t}{t(1 + t^2)^2} \).

Step 1: Identify the Differentiation Rule

To differentiate the function \( h(t) = \frac{\ln t}{1+t^2} \), we will need to use the quotient rule, which is applicable for functions in the form \( \frac{u}{v} \). The rule states that if \( h(t) = \frac{u(t)}{v(t)} \), then \( h'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{(v(t))^2} \).

Step 2: Define the Terms u(t) and v(t)

Assign \( u(t) = \ln t \) and \( v(t) = 1 + t^2 \) for the function \( h(t) = \frac{\ln t}{1+t^2} \).

Step 3: Differentiate u(t) - the Numerator

Differentiate \( u(t) = \ln t \) with respect to \( t \). The derivative is \( u'(t) = \frac{1}{t} \).

Step 4: Differentiate v(t) - the Denominator

Differentiate \( v(t) = 1 + t^2 \). The derivative is \( v'(t) = 2t \).

Step 5: Apply the Quotient Rule

Substitute \( u(t) = \ln t \), \( v(t) = 1 + t^2 \), \( u'(t) = \frac{1}{t} \), and \( v'(t) = 2t \) into the quotient rule formula:\[ h'(t) = \frac{\left(\frac{1}{t}\right)(1 + t^2) - (\ln t)(2t)}{(1 + t^2)^2} \]

Step 6: Simplify the Expression

Calculate and simplify the expression:1. Multiply out: \[ \frac{1}{t}(1 + t^{2}) = \frac{1}{t} + t \]2. Multiply out: \( (\ln t)(2t) = 2t\ln t \)3. Substitute back in:\[ h'(t) = \frac{\frac{1}{t} + t - 2t\ln t}{(1 + t^2)^2} \]Combine terms to simplify the expression.

Step 7: Final Simplified Form of the Derivative

The derivative of the function is:\[ h'(t) = \frac{\frac{1 + t^2}{t} - 2t\ln t}{(1 + t^2)^2} \]Thus, the simplified expression for the derivative is:\[ h'(t) = \frac{1 + t^2 - 2t^2 \ln t}{t(1 + t^2)^2} \]

Key Concepts

Quotient Rule

In calculus, the quotient rule is essential for finding the derivative of functions that can be expressed as a quotient of two other functions. The rule is particularly useful when you have a function written as \( \frac{u(t)}{v(t)} \), where both the numerator \( u(t) \) and the denominator \( v(t) \) are themselves functions of \( t \). The formula for the derivative using the quotient rule is:

• \( h'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{(v(t))^2} \)

Here’s how it works:

• First, differentiate \( u(t) \) to get \( u'(t) \).
• Then, differentiate \( v(t) \) to get \( v'(t) \).
• Substitute these derivatives and the original functions into the quotient rule formula.

This method ensures you handle the derivative of a quotient correctly, avoiding common mistakes like product rule application instead.

Natural Logarithm Differentiation

The natural logarithm function, written as \( \ln x \), has unique properties that make it important in calculus. Differentiating \( \ln t \) with respect to \( t \) is straightforward due to its simple derivative:

• \( \frac{d}{dt} \ln t = \frac{1}{t} \)

Whenever you encounter logarithmic functions in differentiation problems, remember:

• The derivative of \( \ln t \) is always \( \frac{1}{t} \).
• This rule applies unless there are additional transformations or compositions involved, such as a chain rule scenario.

This makes the natural logarithm one of the basic functions you can quickly differentiate directly. It is useful in simplifying the process of finding derivatives involving log expressions.

Calculus Problem Solving

Solving calculus problems like differentiating complex functions can seem daunting, but following a systematic approach makes it manageable. Here are steps that help in tackling derivative problems involving both simple and complex functions:

• First, carefully identify the form of the function you are working with. Determine whether it's straightforward or a combination (like a quotient or product).
• Decide which differentiation rules apply. For our function example, we used the quotient rule due to its \( \frac{\ln t}{1 + t^2} \) structure.
• Carry out differentiation steps methodically, applying the chosen rules to each part of the function.
• Finally, simplify the expression. Simplification ensures you have the cleanest form of a solution, which is crucial for further analysis or applications.

By consistently applying these strategies, you’ll gain confidence and efficiency in solving calculus problems, making complex derivatives feel simpler.