Problem 50 - Calculus and its applications

Question

Find \(y^{\prime}\) $$ \text { If } f(x)=x^{2}+4 x-5, \text { find } f^{\prime}(10) $$

Step-by-Step Solution

Answer

The derivative \(f'(10)\) is 24.

Step 1: Understand the Problem

We need to find the derivative of the function \(f(x) = x^2 + 4x - 5\) at a specific point, which is \(x = 10\). This requires first finding the derivative \(f'(x)\), and then evaluating it at \(x=10\).

Step 2: Find the Derivative of \(f(x)\)

The derivative of a function is found by applying the power rule, which states that if \(f(x) = ax^n\), then \(f'(x) = anx^{n-1}\). For \(f(x) = x^2 + 4x - 5\), the derivative is \(f'(x) = 2x + 4\).

Step 3: Evaluate the Derivative at \(x=10\)

Substitute \(x = 10\) into the derivative \(f'(x) = 2x + 4\). Calculate \(f'(10) = 2(10) + 4 = 20 + 4 = 24\).

Step 4: Conclusion

Thus, the derivative of \(f(x)\) evaluated at \(x = 10\) is \(f'(10) = 24\).

Key Concepts

Understanding the Power Rule

The power rule is a fundamental technique in calculus used to find the derivative of polynomial functions quickly. It states that, given a function in the form of \(f(x) = ax^n\), the derivative of this function is \(f'(x) = nan^{n-1}\). This rule basically helps us determine how the function changes as its input \(x\) changes. For each term in the function, you multiply the coefficient by the exponent, and then decrease the exponent by one.

In our original exercise, the function is \(f(x) = x^2 + 4x - 5\). Let's break this down using the power rule:

• For the term \(x^2\), the derivative is \(2x^{2-1} = 2x\).
• For the term \(4x\), the derivative is simply \(4\) since the exponent of \(x\) is 1, and \(4x^0 = 4\).

Applying the power rule to each part, we find the derivative of the function to be \(f'(x) = 2x + 4\). The constant term \(-5\) disappears in the derivative because the derivative of any constant is zero. This simplifies the process of finding derivatives, especially for functions with many terms.

Function Evaluation Essentials

Once we have the derivative function, the next step is to evaluate the function at a specific point. This is called function evaluation. In calculus, evaluating the derivative at a specific point helps us understand the rate of change or slope of that function right at that point.

In the context of our problem, we need to substitute \(x = 10\) into the derivative function \(f'(x) = 2x + 4\). This involves simply plugging the value of \(x\) into the equation:

• Replace \(x\) with 10 in the equation \(f'(x) = 2x + 4\).
• This gives us \(f'(10) = 2(10) + 4\).

The function evaluation is a straightforward process where you're essentially finding out the exact value of the derivative for a given \(x\). This gives precise information about the behavior of the function at that specific point.

Derivative at a Specific Point

Finding the derivative at a specific point combines your understanding of differentiation and function evaluation. It helps measure how the function behaves precisely at a certain point. In our problem, we've already derived that \(f'(x) = 2x + 4\). To find \(f'(10)\), we evaluate this expression at \(x = 10\).

Let's see how it's done:

• Use the derivative \(f'(x) = 2x + 4\).
• Plug in \(x = 10\) into this derivative to get \(f'(10) = 2(10) + 4\).
• Calculate the expression: \(2(10) = 20\), then add \(4\), which results in \(24\).

Thus, the value of the derivative of the function at \(x = 10\) is \(24\). This means, at \(x = 10\), the function's rate of change is \(24\). This is crucial in understanding how the graph of the function is behaving right at that point, whether it's increasing or decreasing, and at what steepness. The concept can be incredibly useful for interpreting real-world phenomena represented by the function.