Question

In Exercises, find the derivative of the function. $$ y=\ln \frac{x^{2}}{x^{2}+1} $$

Short answer

The derivative of the function is \(\frac{1}{x}\).

Step-by-step solution

Identify the function type

This function is a natural logarithm \(\ln(x)\) applied on a function \(\frac{x^{2}}{x^{2}+1}\). Therefore, to calculate its derivative, the chain rule will need to be applied.

Apply the chain rule

The chain rule states that the derivative of a composition of functions is the product of the derivative of the outer function times the derivative of the inner function. In this case, for \(\ln(u)\), its derivative is \(\frac{1}{u}\). By applying the chain rule, the derivative of \(\ln(\frac{x^{2}}{x^{2}+1})\) is \(\frac{1}{\frac{x^{2}}{x^{2}+1}} \cdot \frac{d}{dx}(\frac{x^{2}}{x^{2}+1})\).

Apply the quotient rule

The derivative of the quotient \(\frac{u}{v}\) is \(\frac{vu'-uv'}{v^{2}}\). Here, \(u = x^{2}\) and \(v = x^{2} + 1\), so the derivative \(\frac{d}{dx}(\frac{x^{2}}{x^{2}+1})\) is \(\frac{(x^{2} + 1)\cdot 2x - x^{2}\cdot 2x}{(x^{2} + 1)^{2}}\).

Simplify the derivative

Simplify the multiplication for the derivative to get the final answer. The final output is \(\frac{2x}{2(x^{2}+1)}\cdot \frac{1}{\frac{x^{2}}{x^{2}+1}}\) which amounts to \(\frac{1}{x}\). This is the derivative of the function \(y=\ln(\frac{x^{2}}{x^{2}+1})\).

Key concepts

Natural Logarithm

The natural logarithm, represented as \( \ln(x) \), is a mathematical function that is the inverse of the exponential function with base \( e \), where \( e \) is approximately 2.718. It's often used to solve for the exponent in equations of the form \( e^y = x \).
When dealing with functions inside a natural logarithm, such as \( \ln(u) \), where \( u \) is another function, it's essential to apply specific rules when finding derivatives.
The derivative of the natural logarithm \( \ln(u) \) is \( \frac{1}{u} \), which indicates how the logarithm function changes with respect to \( u \). This derivative acts as the outer function in expressions where \( u \) is a function of \( x \).
Knowing how to find the derivative of a natural logarithm is crucial because it appears frequently in a wide array of mathematical and real-world problems, including growth models and in the study of compound interest.

Chain Rule

The chain rule is a fundamental principle in calculus used for finding the derivative of composite functions. A composite function is one that can be thought of as \( f(g(x)) \), where \( f \) is the outer function and \( g(x) \) is the inner function.
To find the derivative of such functions, the chain rule tells us to take the derivative of the outer function evaluated at the inner function and multiply it by the derivative of the inner function. In mathematical terms, if you have \( y = f(g(x)) \), its derivative \( \frac{dy}{dx} \) is \( f'(g(x)) \cdot g'(x) \).
In the context of our problem, we apply the chain rule to \( y = \ln \left( \frac{x^2}{x^2 + 1} \right) \). Here, the outer function is \( \ln(u) \) and the inner function is \( \frac{x^2}{x^2 + 1} \). So, the chain rule helps by first finding the derivative of \( \ln(u) \) which is \( \frac{1}{u} \), then multiplying by the derivative of the inner expression.

Quotient Rule

The quotient rule is a technique in calculus for finding the derivative of a function that is the quotient of two other functions. If you have a function in the form \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the quotient rule states:

• The derivative \( \frac{d}{dx}\left(\frac{u}{v}\right) \) is given by \( \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \).
• This formula works because it effectively combines the product rule and the idea of derivatives as rates of change.

In our specific problem, we calculate the derivative of \( \frac{x^2}{x^2+1} \). Here, \( u = x^2 \) and \( v = x^2 + 1 \). Thus, the derivatives are \( u' = 2x \) and \( v' = 2x \).
Plugging these into the quotient rule formula, we get \( \frac{(x^2 + 1) \cdot 2x - x^2 \cdot 2x}{(x^2 + 1)^2} \). This derivative is key to solving the larger problem when applied alongside the chain rule and natural logarithm rules.