Problem 56 - Calculus

Question

Find the derivatives in algebraically.$$f(x)=5 x^{2} \text { at } x=10$$

Step-by-Step Solution

Answer

The derivative of \( f(x) = 5x^2 \) at \( x = 10 \) is 100.

Step 1: Identify the Function

The given function is \( f(x) = 5x^2 \). This is a quadratic function, and we're tasked with finding its derivative.

Step 2: Apply the Power Rule

Using the power rule for derivatives, \( \frac{d}{dx} [x^n] = nx^{n-1} \), we differentiate \( f(x) = 5x^2 \). The derivative is \( f'(x) = 2 \cdot 5x^{2-1} = 10x \).

Step 3: Evaluate the Derivative at a Specific Point

The problem asks us to find the derivative at \( x = 10 \). Substitute \( x = 10 \) into the derivative function \( f'(x) = 10x \), which gives \( f'(10) = 10 \cdot 10 = 100 \).

Key Concepts

Power Rule

When dealing with derivatives, the power rule is one of the most basic tools to use. It is particularly useful for differentiating polynomial functions. The power rule states: \( \frac{d}{dx} [x^n] = nx^{n-1} \). In simple terms, it allows you to take the exponent of the variable, multiply it by the coefficient, and then reduce the power by one.

By applying the power rule, we greatly simplify the process of differentiation.

• Identify the highest power of the variable \( x \) in the term you want to differentiate.
• Multiply this power by the coefficient of the term.
• Subtract one from the original power.

This results in the derivative for that specific term. In our example of finding the derivative of \( f(x) = 5x^2 \), the power was 2, so we multiply 2 by the coefficient 5 to get \( 10x^{1} \), or simply \( 10x \).
Understanding and mastering this rule can make derivative calculations fast and efficient, especially when dealing with more complex polynomial expressions.

Quadratic Function

Quadratic functions form the foundation for many concepts in calculus and algebra. They are typically expressed in the general form \( ax^2 + bx + c \). For the purpose of our problem, we are looking at \( f(x) = 5x^2 \), a simple quadratic function where \( a = 5 \) and both \( b \) and \( c \) are 0.

Quadratic functions are recognized by their distinctive parabolic graphs, which open upwards if \( a > 0 \) and downwards if \( a < 0 \). These functions have a unique characteristic where the highest exponent of the variable is 2. When you find the derivative of a quadratic function, you'll notice that the process turns it into a linear function. As seen, upon differentiating \( 5x^2 \), we get \( 10x \). This linearity can be extremely useful for determining the slope or rate of change at any point along the curve.

Quadratic functions are pervasive in modeling real-world scenarios like projectile motion and cost functions. Grasping how to differentiate them unlocks the ability to analyze the behaviors of these scenarios effectively.

Evaluate Derivative

After deriving a function, the next practical step is to evaluate the derivative at a specific point. This involves substituting the 'x' value of interest into the derivative to find the corresponding rate of change or slope of the tangent at that point.

For the function \( f(x) = 5x^2 \), we calculated the derivative to be \( f'(x) = 10x \). To evaluate this derivative at \( x = 10 \), replace \( x \) with 10 in the expression: \( f'(10) = 10 \times 10 = 100 \).

• Substitute the given \( x\) value into the derivative function.
• Simplify the result to determine the slope or rate of change at that specific point.

So in this problem, \( f'(10) = 100 \) tells us that the slope of the tangent to the curve at \( x = 10 \) is 100. In practical terms, this means that as \( x \) changes near 10, the value of \( f(x) \) increases by 100 for every unit increase in \( x \). This step is crucial for understanding how a function behaves at specific points, allowing for predictions and insights in varied applications.