Problem 16 - Precalculus: Mathematical for Calculus

Question

Find the derivative of the function at the given number. $$g(x)=2 x^{2}+x^{3} \quad \text { at } 1$$

Step-by-Step Solution

Answer

The derivative of the function at 1 is 7.

Step 1: Differentiate each term of the function

Start by finding the derivative of each term in the function separately. The function given is \( g(x) = 2x^2 + x^3 \). Use the power rule: \( \frac{d}{dx}x^n = nx^{n-1} \).

Step 2: Apply the power rule to each term

For the term \( 2x^2 \), the derivative is \( 2 \times 2x^{2-1} = 4x \). For the term \( x^3 \), the derivative is \( 3x^{3-1} = 3x^2 \).

Step 3: Combine the derivatives

Now, combine the derivatives of all terms to get the derivative of the function. The derivative of \( g(x) \) is \( g'(x) = 4x + 3x^2 \).

Step 4: Evaluate the derivative at the given number

Substitute \( x = 1 \) into the derivative \( g'(x) = 4x + 3x^2 \). This gives \( g'(1) = 4 \times 1 + 3 \times 1^2 = 4 + 3 = 7 \).

Key Concepts

Derivative

A derivative in calculus represents how a function changes as its input changes. It is often described as the "rate of change" or "slope" of the function at a particular point. Imagine you're on a roller coaster and you want to see how steep the track is at different spots. Derivatives do just that for functions.

In mathematical terms, the derivative of a function \( f(x) \) at a point \( x \) is the slope of the tangent line to the graph of the function at that point. This means that if you zoom into the curve of the function very closely, the curve starts looking like a straight line. The slope of this line is the derivative.

When solving derivative problems, like differentiating \( g(x) = 2x^2 + x^3 \), you're typically trying to find how fast \( g(x) \) is changing at any given \( x \)-value. This is useful for understanding the behavior of a function, optimizations, and more nuanced calculus applications.

Power Rule

The power rule is a quick shortcut in calculus for finding the derivative of a function. It's particularly useful for polynomials, like the function \( g(x) = 2x^2 + x^3 \).

Here's how the power rule works: If you have a term of the form \( x^n \), the derivative is \( nx^{n-1} \).

• You multiply the exponent \( n \) by the coefficient in front of the term.
• Then, you decrease the exponent by 1.

This rule saves time compared to the more formal definition of a derivative. For instance:

• For \( x^3 \), the power rule gives us a derivative of \( 3 imes x^{3-1} = 3x^2 \).
• For the term \( 2x^2 \), applying the power rule results in \( 2 \times 2x^{2-1} = 4x \).

By quickly applying this rule to each term in a function, you can derive the overall function efficiently.

Differentiation

Differentiation is the process by which we find a function's derivative. It's a fundamental tool in calculus, used to understand how functions behave.

When you differentiate, you consider each term of your function individually, applying rules to transform it into its derivative. Our earlier example \( g(x) = 2x^2 + x^3 \) was differentiated term by term:

• The first term, \( 2x^2 \), becomes \( 4x \) using the power rule.
• The second term, \( x^3 \), becomes \( 3x^2 \).

Once each term has been differentiated, they are combined to give the complete derivative of the function. In this case, \( g'(x) = 4x + 3x^2 \).

Finally, to find the value at specific points, such as \( x = 1 \), substitute \( x \) into the derivative. This gives \( g'(1) = 4 \times 1 + 3 \times 1^2 = 7 \), showing how steep the function is at that point.