Problem 5 - Applied Calculus

Question

Differentiate the functions in Problems 1-28. Assume that \(A\), \(B\), and \(C\) are constants. \(y=2^{x}+\frac{2}{x^{3}}\)

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = 2^x \ln(2) - 6x^{-4} \).

Step 1: Differentiate the first term

The function is given as \(y = 2^x + \frac{2}{x^3}\). The derivative of the exponential function \(2^x\) involves using the natural logarithm, so \(\frac{d}{dx}[2^x] = 2^x \ln(2)\).

Step 2: Differentiate the second term using the power rule

The second term to differentiate is \(\frac{2}{x^3}\). Rewrite it as \(2x^{-3}\). Using the power rule \(\frac{d}{dx}[x^n] = nx^{n-1}\), find that \(\frac{d}{dx}[2x^{-3}] = 2(-3)x^{-4} = -6x^{-4}\).

Step 3: Combine the derivatives

Now combine the derivatives calculated in the previous steps. This gives: \( \frac{dy}{dx} = 2^x \ln(2) - 6x^{-4}\).

Key Concepts

Exponential Function Differentiation

When you're asked to differentiate exponential functions like \(2^x\), it's a little different from differentiating polynomial expressions. The unique aspect of exponential functions is that the base is a constant while the exponent contains the variable. To approach such differentiation, we use the natural logarithm. So, for the function \(2^x\), the derivative is calculated using the formula:

• \( \frac{d}{dx}[a^x] = a^x \ln(a) \)

For \(2^x\), you apply this by plugging in 2 where \(a\) stands for the base, resulting in:

• \( \frac{d}{dx}[2^x] = 2^x \ln(2) \)

This gives you the rate at which the function \(2^x\) changes with respect to \(x\). Differentiation of exponential functions is crucial as they often model growth or decay processes in real-world scenarios.

Power Rule

The power rule is an essential tool in calculus, simplifying the process of differentiating expressions where variables are raised to a power. This rule states that if you have a function \(x^n\), its derivative is:

• \( \frac{d}{dx}[x^n] = nx^{n-1} \)

In our problem, you face the term \( \frac{2}{x^3} \), which can be rewritten using negative exponents as \(2x^{-3}\). Applying the power rule here means multiplying the exponent by the coefficient and reducing the exponent by one:

• \( \frac{d}{dx}[2x^{-3}] = 2(-3)x^{-4} = -6x^{-4} \)

This technique makes it easy to handle more complicated functions. It reduces the complexity of working with polynomials, especially those with negative or fractional exponents.

Calculus Problem-Solving

When tackling problems in calculus, especially derivatives, a systematic approach helps manage each component of the function separately. Here's how you can effectively solve such problems:
First, break down the function into individual parts or terms. Each term might require a different rule:

• Use the exponential function differentiation techniques for exponential terms.
• Apply the power rule for power or polynomial expressions.

For example, in differentiating \(y = 2^x + \frac{2}{x^3}\), identify how each term is distinct and requires different methods for differentiation. After calculating the derivative for each part:

• Combine them into a single expression.

Our solution becomes \( \frac{dy}{dx} = 2^x \ln(2) - 6x^{-4} \).
Clearly organizing your work can lead to successful problem-solving in calculus, allowing you to address each component accurately and efficiently.