Problem 30 - Calculus and its applications

Question

Differentiate each function. \(F(x)=(x+3)^{2}\) \(\left[\right.\) Hint: \(\left.(x+3)^{2}=(x+3)(x+3) .\right]\)

Step-by-Step Solution

Answer

The derivative of \(F(x) = (x+3)^2\) is \(F'(x) = 2x + 6\).

Step 1: Expand the function

Using the hint provided, expand the function: \[(x+3)^2 = (x+3)(x+3) = x^2 + 3x + 3x + 9 = x^2 + 6x + 9.\] This step simplifies the expression, making it easier to differentiate.

Step 2: Differentiate using the Power Rule

Apply the power rule for derivatives, which states that \( \frac{d}{dx}(x^n) = nx^{n-1} \). Differentiate each term separately: - The derivative of \(x^2\) is \(2x\). - The derivative of \(6x\) is \(6\). - The derivative of the constant \(9\) is \(0\). This gives us \(2x + 6 + 0\).

Step 3: Write the final derivative

Combine the results from the differentiated terms to get the final expression: \[F'(x) = 2x + 6.\] This is the derivative of the function \(F(x)=(x+3)^2.\)

Key Concepts

Power Rule

The Power Rule is a fundamental concept in differentiation, which simplifies the process of finding derivatives for polynomial terms. When you have a term like \(x^n\), applying the Power Rule involves bringing down the exponent as a coefficient and then reducing the exponent by one. Mathematically, this is expressed as:

• \( \frac{d}{dx}(x^n) = nx^{n-1} \)

This means:

• For \(x^2\), the derivative is \(2x^{2-1} = 2x\).
• If you had \(x^3\), it would become \(3x^{3-1} = 3x^2\).
• With constant terms like \(x^0\) (which is just a constant, like \(9\) in our problem), the derivative is zero as constants don't change.

Using the Power Rule is especially helpful because it provides a straightforward, consistent method for differentiating polynomial terms, making it essential for calculus students to master.

Polynomial Functions

Polynomial functions are expressions that involve sums of powers of \(x\) with constant coefficients. These functions are significant in algebra and calculus due to their relatively simple structure and the ease with which they can be analyzed and differentiated. A general polynomial function looks like:

• \(f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0\)

In our exercise, \((x+3)^2\) can be expanded into a polynomial:

• \(x^2 + 6x + 9\)

This makes it easier to apply differentiation rules because each term is isolated. Polynomial functions are important because differentiating them helps us find rates of change and slopes of tangent lines among other things. They serve as building blocks for more complex expressions in calculus.

Step-by-Step Solution

Approaching problems with a clear, step-by-step method simplifies complex tasks like differentiation. First, by using hints or provided information, break down expressions into simpler parts. For instance, to differentiate \((x+3)^2\), we:

• Expanded it to \(x^2 + 6x + 9\) making each term manageable.

Next, apply the Power Rule to each term in the expanded polynomial. Each derivative is assessed:

• \(x^2\) becomes \(2x\) due to the Power Rule.
• \(6x\) simplifies to \(6\) since the derivative of \(x\) is 1.
• The constant \(9\) has a derivative of \(0\).

Finally, combine these results to write the complete derivative expression, \(F'(x) = 2x + 6\). This systematic approach not only ensures accuracy but also builds a strong foundation for dealing with more elaborate functions in calculus.