Problem 70 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to the given independent variable. $$y=\log _{25} e^{x}-\log _{5} \sqrt{x}$$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \frac{1}{\ln 25} - \frac{1}{2x\ln 5}\)

Step 1: Apply Logarithm Change of Base Formula

The given function involves logarithms with different bases. Start by converting these logarithms to base 10 or natural logarithms (base \(e\)). Use the change of base formula: \[\log_a b = \frac{\log_c b}{\log_c a}\].For \(\log_{25} e^x\) and \(-\log_{5} \sqrt{x}\), apply the change of base formula using natural logarithms:\[y = \frac{\ln e^x}{\ln 25} - \frac{\ln \sqrt{x}}{\ln 5}\].

Step 2: Simplify the Function

Simplify each term using properties of logarithms. Recall that \(\ln e^x = x\) and \(\ln \sqrt{x} = \frac{1}{2} \ln x\). Thus, rewrite \(y\) as:\[y = \frac{x}{\ln 25} - \frac{1}{2 \ln 5} \ln x\].

Step 3: Differentiate with Respect to x

Now differentiate the simplified expression of \(y\) with respect to \(x\). Use the differentiation rules for \(x\) and \(\ln x\):The first term differentiation: \[\frac{d}{dx} \left( \frac{x}{\ln 25} \right) = \frac{1}{\ln 25}\].The second term differentiation involves the chain rule:\[\frac{d}{dx} \left( \frac{1}{2\ln 5} \ln x \right) = \frac{1}{2\ln 5} \cdot \frac{1}{x} = \frac{1}{2x\ln 5}\].

Step 4: Combine Derivatives

Combine the results from the differentiation of each term to find \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{1}{\ln 25} - \frac{1}{2x\ln 5}\].

Key Concepts

Logarithms

Logarithms are the inverse operations to exponentiation, meaning they help us find the power to which a number must be raised to obtain another number. They are commonly expressed in a mathematical form as \(log_b a\), which reads as "log of \(a\) to the base \(b\)." In our exercise, base 25 and base 5 logarithms are used, which can be tricky to handle directly in differentiation. Hence, we convert them using more familiar bases, like base 10 or natural base \(e\).

• In computational terms, logarithms help simplify multiplications into additions, making calculations easier.
• They are useful in solving equations involving exponential growth and help in examining the behavior of curves in calculus.

To deal with different bases in this problem, we apply the change of base formula, which is a simple yet powerful tool.

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions. It is essential when differentiating functions where one function is nested inside another. For example, in our exercise, within the expression \( \ln \sqrt{x} \), the function \(\sqrt{x}\) itself is a composition of \(x^{1/2}\) and other operations handled by the natural log \(\ln\).

• The chain rule states that the derivative of a composite function \(f(g(x))\) is \(f'(g(x)) \cdot g'(x)\).
• It helps break down complex differentiations into manageable steps.

In our solution, we apply the chain rule when differentiating terms like \(\ln \sqrt{x}\), effectively simplifying a complex differentiation process into easier parts.

Change of Base Formula

The change of base formula allows us to convert a logarithm with any base to a more common base like 10 or \(e\), which are simpler to handle, especially in differentiation and integration. The formula is expressed as:\[log_a b = \frac{log_c b}{log_c a}\]Using this formula:

• We can transform \(\log_{25} e^x\) to a function in terms of natural logs, \(\ln\).
• Similarly, \(\log_{5} \sqrt{x}\) is converted in a similar fashion.

By converting these logarithms to natural base \(e\), they become easier to differentiate because the rules for \(\ln\) functions are much simpler. This simplification is at the heart of successfully solving the given problem and finding the derivative.

Differentiation Rules

Differentiation rules provide a set framework for deriving functions. For straightforward functions like powers of \(x\) or exponential functions, these rules let us handle more complicated expressions systematically. In our exercise, we apply basic differentiation rules to each simplest function derived from the original expression.

• The derivative of a constant multiple of a function, for instance, \(\frac{d}{dx}(c\cdot f(x))\), is \(c\cdot f'(x)\).
• For \(\ln x\), the derivative with respect to \(x\) is \(\frac{1}{x}\).

In finding \(\frac{dy}{dx}\), we use these rules step by step to individually differentiate each part of our simplified expression. This methodical application ensures an accurate overall derivative.