Question

Find \(f^{\prime}(x)\) for each function. $$ f(x)=x^{\pi} \cdot \pi^{x} $$

Short answer

The derivative is \(f'(x) = \pi^x (\pi x^{\pi-1} + x^{\pi} \ln \pi)\).

Step-by-step solution

Understand the Problem

We need to find the derivative of the function \(f(x) = x^{\pi} \cdot \pi^{x}\). This is a product of two functions: \(u(x) = x^{\pi}\) and \(v(x) = \pi^{x}\). To find the derivative, we will apply the product rule for differentiation.

Identify Function Components

The function is \(f(x) = x^{\pi} \cdot \pi^{x}\). Here, \(u(x) = x^{\pi}\) and \(v(x) = \pi^{x}\). We will differentiate each component separately.

Differentiate \(u(x) = x^{\pi}\)

Differentiate \(u(x) = x^{\pi}\) using the power rule. The derivative is \(u'(x) = \pi \cdot x^{\pi - 1}\).

Differentiate \(v(x) = \pi^{x}\)

The derivative of \(v(x) = \pi^{x}\) is found using the exponential rule. The derivative is \(v'(x) = \ln(\pi) \cdot \pi^{x}\).

Apply Product Rule

The product rule states that if \(f(x) = u(x) \cdot v(x)\), then \(f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)\). Substitute the derivatives calculated:\[f'(x) = \pi \cdot x^{\pi - 1} \cdot \pi^{x} + x^{\pi} \cdot \ln(\pi) \cdot \pi^{x}\]

Simplify the Expression

Factor out the common term \(\pi^{x}\) to get:\[f'(x) = \pi^{x} \cdot \left( \pi \cdot x^{\pi - 1} + x^{\pi} \cdot \ln(\pi) \right)\] This is the derivative \(f^{\prime}(x)\).

Key concepts

Product Rule

When dealing with functions that are a product of two separate functions, the product rule is your friend. It's a fundamental rule in calculus used whenever you differentiate functions written in the form of a product. For instance, if you have a function given by the product of two functions, say \( f(x) = u(x) \cdot v(x) \), the product rule comes into play.
To differentiate such a function, the product rule states you need to calculate the derivative as follows:

• First, find the derivative of the first function \( u(x) \), which we call \( u'(x) \).
• Next, keep the second function \( v(x) \) as is.
• Add to this the derivative of the second function \( v'(x) \) while keeping the first function \( u(x) \) unchanged.

Using this rule, the derivative \( f'(x) \) can be expressed as: \[ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \] Understanding this rule is essential for calculus students, as it makes the process of differentiating products straightforward and logical.

Power Rule

The power rule is one of the simplest and most widely used rules in differentiation, particularly useful for handling functions in the form \( x^n \), where \( n \) is any real number. It's a quick way to find the derivative without much fuss. The power rule states that if you have a function \( x^n \), the derivative is given by pulling the exponent down in front and reducing the exponent by one:
\[ \frac{d}{dx} x^n = n \cdot x^{n-1} \]
So, for example, if you need to differentiate \( u(x) = x^{\pi} \), using the power rule, the derivative becomes \( u'(x) = \pi \cdot x^{\pi - 1} \).
This rule applies to any term with a power, making it a valuable tool for calculus students tackling a wide range of problems.

Exponential Differentiation

Differentiating exponential functions often requires a different approach compared to polynomials. Standard exponential functions take the form \( a^x \) for some constant \( a \). The derivative of an exponential function, like \( v(x) = \pi^x \), involves a key constant: the natural logarithm of the base, \( \ln(a) \).
The general rule of thumb for differentiating such functions is:
\[ \frac{d}{dx} a^x = \ln(a) \cdot a^x \] For instance, for \( v(x) = \pi^x \), the derivative is \( v'(x) = \ln(\pi) \cdot \pi^x \). This differentiation rule highlights how logarithms and exponentials are interconnected.
Understanding this method is crucial as exponential functions often appear in various scientific and mathematical contexts, from compound interest calculations to population growth models.