Problem 79 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to the given independent variable. $$y=3^{\log _{2} t}$$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dt} = \log_2 3 \cdot t^{\log_2 3 - 1} \).

Step 1: Identify the function form

The function given is \( y = 3^{\log_2 t} \). This is an exponential function where the exponent is a logarithmic function of \( t \).

Step 2: Transform the function

To simplify taking the derivative, rewrite the function using properties of logarithms and exponents. Use the identity \( a^{\log_b x} = x^{\log_b a} \), therefore rewriting \( y \) as \( y = t^{\log_2 3} \).

Step 3: Differentiate using the power rule

Now that the function is \( y = t^{\log_2 3} \), we can differentiate using the power rule for differentiation. The derivative of \( t^n \) with respect to \( t \) is \( nt^{n-1} \), thus: \[ \frac{dy}{dt} = \log_2 3 \cdot t^{\log_2 3 - 1} \].

Step 4: Simplify the derivative

The derivative \( \frac{dy}{dt} = \log_2 3 \cdot t^{\log_2 3 - 1} \) is already in its simplest form, expressing the rate of change of \( y \) with respect to \( t \).

Key Concepts

Logarithmic functions

Logarithmic functions are a fundamental concept in mathematics, often represented as \(\log_b x\), where \(b\) is the base and \(x\) is the argument. The logarithm answers the question: "To what power must we raise \(b\) to obtain \(x\)?" This concept is critical when working with exponential functions, as logarithms serve as the inverse.
For example, if \(b^y = x\), then \(\log_b x = y\). In this exercise, we encountered \(\log_2 t\), representing the power to which 2 must be raised to get \(t\).

One useful property of logarithms is the change of base formula: \(\log_b x = \frac{\log_c x}{\log_c b}\). This property can simplify calculations, especially when using calculators that only support certain bases. Furthermore, understanding this property is crucial for manipulating expressions involving logarithms.

Exponential functions

Exponential functions are equations in which a variable appears as an exponent. They have a form like \(a^x\), where \(a\) is a constant and \(x\) is the exponent. These functions are known for their rapid growth or decay, depending on whether the base \(a\) is greater than or less than 1.
In this exercise, the exponential function is \(3^{\log_2 t}\). To find its derivative, we transformed it using the property \(a^{\log_b x} = x^{\log_b a}\). This transformation is essential as it converts a more complex-looking function into a power of \(t\), which simplifies differentiation.

Exponential functions have real-world applications, such as population growth models and radioactive decay. Understanding the behavior of these functions is important in both theoretical and applied contexts.

Power rule for differentiation

The power rule is a basic rule for finding the differentiation of functions with the form \(x^n\). The rule states that if \(y = x^n\), then \(\frac{dy}{dx} = nx^{n-1}\). This rule is straightforward and speeds up the differentiation process tremendously.
In our exercise, we used the power rule to differentiate \(t^{\log_2 3}\). Here, \(n\) is \(\log_2 3\) and the power rule applies directly, resulting in \(\frac{dy}{dt} = \log_2 3 \cdot t^{\log_2 3 - 1}\).

The power rule is not only useful for powers of a variable but also serves as the basis for more complex differentiation techniques, including the chain rule and the product rule. Mastering these can greatly improve one's ability to tackle calculus problems.