Problem 56 - Calculus of a Single Variable

Question

Finding a Derivative In Exercises \(37-58\) , find the derivative of the function. (Hint: In some exercises, you may find it helpful to apply logarithmic properties before differentiating.) $$ g(x)=\log _{5} \frac{4}{x^{2} \sqrt{1-x}} $$

Step-by-Step Solution

Answer

\(- \frac{2}{x \ln(5) } + \frac{1}{2(1-x) \ln(5) } \)

Step 1: Apply Logarithmic Properties

To simplify the expression in the logarithm, we will use two properties of logarithms. The property \( \log_b(a/c) = \log_b(a) - \log_b(c) \) is applicable here, and also, the property \( \log_b(\sqrt(a)) = \frac{1}{2} \log_b(a) \). Therefore, we can rewrite the function as following: \( g(x) = \log_5(4) - \log_5(x^2) - \frac{1}{2} \log_5(1-x) \).

Step 2: Differentiate using derivative logarithm rules and chain rule

The derivative of \( \log_5(a) \) is \( \frac{1}{a \ln(5)} \), we also need to apply chain rule for \( x^2 \) and \( 1-x \). Therefore, the derivative of the function \( g(x) \) is as follows: \( g'(x) = 0 - \frac{2x}{x^2 \ln(5) } - \frac{1}{2(1-x) \ln(5)} \cdot (-1) = - \frac{2}{x \ln(5) } + \frac{1}{2(1-x) \ln(5) } \).

Step 3: Simplify the Result

Now simplify the expression from the last step to get a shorter result. Thus, the answer is \( g'(x) = - \frac{2}{x \ln(5) } + \frac{1}{2(1-x) \ln(5) } \).

Key Concepts

Logarithmic Properties

Understanding logarithmic properties is essential when dealing with logarithmic functions, especially before differentiating them. Logarithms, by definition, are the inverses of exponential functions and have properties that make them simpler to work with in various mathematical contexts.

First and foremost, one of the key properties often used is the quotient rule: for any positive numbers a, b, and base c, the logarithm of a quotient is the difference of the logarithms, represented as \[\begin{equation}\log_c\left(\frac{a}{b}\right) = \log_c(a) - \log_c(b).\end{equation}\]Another crucial property is the power rule, which states that the logarithm of a power is the exponent times the logarithm of the base:\[\begin{equation}\log_c(a^r) = r \log_c(a).\end{equation}\]Also useful is the understanding of how logarithms behave with roots, like square roots: the logarithm of a square root is half the logarithm of the argument. This gives us\[\begin{equation}\log_c(\sqrt{a}) = \frac{1}{2} \log_c(a).\end{equation}\]By applying these properties, we can simplify complex logarithmic expressions into smaller, more manageable pieces before applying calculus tools, such as differentiation.

Chain Rule in Calculus

When you encounter a composite function, one that is made up of two or more functions nested within one another, the chain rule is your go-to tool for differentiation. In simple terms, if you have a function \[\begin{equation} h(x) = f(g(x)),\end{equation}\]then the derivative of h is the derivative of f with respect to g, multiplied by the derivative of g with respect to x. This can be represented as\[\begin{equation} h'(x) = f'(g(x)) \cdot g'(x).\end{equation}\]The chain rule allows us to unravel the layers of composite functions, differentiating from the outside in. When you have a logarithmic function that involves other functions, such as polynomials or radicals, you'll often need to employ the chain rule in order to take the derivative correctly.

Derivative of Logarithmic Functions

Now let's specifically look at the derivative of logarithmic functions. A logarithmic function's general form is \[\begin{equation} f(x) = \log_b(x),\end{equation}\]where b is the base. The derivative of this function with respect to x is\[\begin{equation} f'(x) = \frac{1}{x \ln(b)},\end{equation}\]where \(\ln(b)\) denotes the natural logarithm of the base b. This formula arises from the definition of the natural logarithm function and its integral properties.

In practice, when you differentiate logarithmic functions, you will invariably use this formula. However, you must not forget to apply the chain rule when the argument of the logarithmic function is more complex than a simple x. Keeping the structure of the rule firmly in mind will help you differentiate more intricate logarithmic functions smoothly and accurately.