Problem 25 - Calculus and its applications

Question

Differentiate. $$ y=5^{x} \cdot \log _{2} x $$

Step-by-Step Solution

Answer

The derivative is \( y' = 5^x \ln 5 \cdot \log_2 x + \frac{5^x}{x \ln 2} \).

Step 1: Identify the Rule to Apply

The function to differentiate is a product of two functions: \( y = 5^x \cdot \log_2 x \). This requires the use of the product rule for differentiation, which states \((u imes v)' = u'v + uv'\). In this case, let \( u = 5^x \) and \( v = \log_2 x \).

Step 2: Differentiate \( u = 5^x \)

To differentiate \( u = 5^x \), use the rule \( \frac{d}{dx} a^x = a^x \ln a \). Thus, \( u' = \frac{d}{dx}(5^x) = 5^x \ln 5 \).

Step 3: Differentiate \( v = \log_2 x \)

To differentiate \( v = \log_2 x \), convert it to natural logarithms: \( \log_2 x = \frac{\ln x}{\ln 2} \). Therefore, the derivative is \( v' = \frac{1}{x \ln 2} \), using the derivative of \( \ln x \) which is \( \frac{1}{x} \).

Step 4: Apply the Product Rule

Using the product rule formula, substitute the derivatives: \( y' = u'v + uv' \). Thus, substitute the expressions derived earlier: \( y' = (5^x \ln 5) \cdot \log_2 x + 5^x \cdot \frac{1}{x \ln 2} \).

Step 5: Simplify the Result

The derivative of the given function is \( y' = 5^x \ln 5 \cdot \log_2 x + \frac{5^x}{x \ln 2} \). This is the simplified form of the derivative.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus. It involves finding the derivative of a function, which is a new function that gives the rate of change of the original function with respect to its variable. In simpler terms, differentiation can be seen as calculating how much a function changes as its input changes.

The most basic rule of differentiation is the power rule. If you have a function like \(x^n\), its derivative is \(nx^{n-1}\). With this, you can quickly find the derivative of polynomial functions.

There are many other rules in differentiation, such as the product rule, quotient rule, and chain rule. The product rule, used in our exercise, is very helpful when differentiating a product of two functions. This rule states that if you have two functions multiplied together, \( u \times v \), the derivative is given by \((u'v + uv')\). This allows you to differentiate complex combinations of functions easily.
Understanding these rules and how they are applied is crucial for accurately finding how functions behave as they change.

Exponential Functions

Exponential functions are unique mathematical functions where the variable is in the exponent. A common type is \(a^x\), where \(a\) is a constant base. These functions grow rapidly, much faster than linear or polynomial functions.

Exponential functions are essential across various fields such as biology, physics, and economics because they model growth processes, such as population growth or compound interest. For example, \(5^x\) in our problem represents an exponential function with a base of 5.

Differentiating exponential functions follows a straightforward rule. If the function is \(a^x\), then its derivative is \(a^x \ln a\). Here, \(\ln a\) is the natural logarithm of \(a\). It's crucial to remember this rule when encountering such functions, as it makes differentiating them simple and fast.
Apply this rule confidently whenever you need to work with exponential growth or decay in mathematical equations.

Logarithmic Functions

Logarithmic functions involve the logarithm operation, which is the inverse of exponentiation. For example, \( \log_{2}x \) answers the question "to what power must 2 be raised, to yield \(x\)?" Logarithms come in different bases, but the natural logarithm, \( \ln x \), is a common one, based on the constant \(e\).

Logarithmic functions are beneficial for solving equations where the variable is an exponent. They are also useful for compressing large ranges of numbers and are used in many scientific settings.

For differentiation, the logarithmic derivative’s basic form for a natural logarithm, \(\ln x\), is \(\frac{1}{x}\). When dealing with different bases, such as \(\log_{2}x\), conversion to a natural logarithm is necessary, where \(\log_{b}x = \frac{\ln x}{\ln b}\). This enables easy differentiation using the natural logarithm's derivative rule.
Logarithmic differentiation is powerful, especially when handling products or quotients of functions with slightly more complexity than simple polynomials.