91Ó°ÊÓ

Logarithmic differentiation is a technique used to differentiate functions that involve complex expressions, especially products, quotients, or powers.
By applying logarithms to both sides of a function, we can simplify the differentiation process.
In the given exercise, we use the natural logarithm to turn the complicated fraction into a simpler difference of two logarithms:
\( f(x) = \ln \left( \frac{1 + \sqrt{x}}{1 - \sqrt{x}} \right) = \ln (1 + \sqrt{x}) - \ln (1 - \sqrt{x}) \).
This simplification allows us to use basic logarithm properties and makes it easier to apply differentiation rules.

The chain rule is a fundamental technique in calculus used to differentiate composite functions.
It states that if a function \( y \) can be written as \( y = g(u) \), where \( u \) is another function, \( u = h(x) \), the derivative of \( y \) with respect to \( x \) is given by \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).
In the solution provided, we use the chain rule to differentiate \( g(x) = \ln (1 + \sqrt{x}) \) and \( h(x) = \ln (1 - \sqrt{x}) \).
This involves first finding the derivative of the outer function (the natural logarithm function) and then multiplying it by the derivative of the inner function (inside the parentheses).
For example, differentiating \( g(x) \) we get:
\( g'(x) = \frac{1}{1 + \sqrt{x}} \cdot \frac{d}{dx}(1 + \sqrt{x}) = \frac{1}{1 + \sqrt{x}} \cdot \frac{1}{2 \sqrt{x}} = \frac{1}{2 \sqrt{x}(1 + \sqrt{x})} \).

Calculus is an essential tool used in the life sciences to model and solve various biological problems.
Differential calculus helps in understanding rates of change, such as how a population grows or how a drug concentration decreases in the bloodstream over time.
Logarithmic differentiation and the chain rule allow biologists to tackle complex problems involving growth models, decay rates, and more.
In the context of our exercise, such techniques can be used to study how certain biological quantities behave under specific conditions.

Combining derivatives involves integrating multiple differentiation results to find the overall derivative of a function.
In the given example, after differentiating both \( \ln (1 + \sqrt{x}) \) and \( \ln (1 - \sqrt{x}) \), we combine them to get the final result.
The derivatives are: \( g'(x) = \frac{1}{2 \sqrt{x}(1 + \sqrt{x})} \) and \( h'(x) = \frac{-1}{2 \sqrt{x}(1 - \sqrt{x})} \).
By subtracting these, we obtain:
\( f'(x) = \frac{1}{2 \sqrt{x}(1 + \sqrt{x})} - \frac{-1}{2 \sqrt{x}(1 - \sqrt{x})} \).
To combine these fractions, we find a common denominator and simplify: \( f'(x) = \frac{1}{2 \sqrt{x}(1 + \sqrt{x})(1 - \sqrt{x})} + \frac{1}{2 \sqrt{x}(1 + \sqrt{x})(1 - \sqrt{x})} = \frac{2}{2 \sqrt{x}(1 - x)} = \frac{1}{x(1 - x)} \).
Thus offering the final derivative of the function.