Problem 33 - Calculus - AP Edition

Question

Find \(d y /\left.d x\right|_{x=1},\) given that \(y=1-x^{2}\)

Step-by-Step Solution

Answer

\(-2\)

Step 1: Understand the Problem

We need to find the derivative of the function \( y = 1 - x^2 \) and then evaluate it at \( x = 1 \). This means we ought to find \( \frac{dy}{dx} \) and then substitute \( x = 1 \) to get the final value.

Step 2: Differentiate the Function

To differentiate the function \( y = 1 - x^2 \), we apply basic rules of differentiation. The derivative of a constant is zero, and the derivative of \( -x^2 \) is \( -2x \). Thus, the derivative \( \frac{dy}{dx} \) is \( -2x \).

Step 3: Substitute and Evaluate at x = 1

Now that we have \( \frac{dy}{dx} = -2x \), we substitute \( x = 1 \) into this expression. This gives \( \frac{dy}{dx} = -2(1) = -2 \).

Step 4: Conclusion

After substituting \( x = 1 \), we find that the derivative of the function at this point is \( -2 \). Therefore, \( \left.\frac{dy}{dx}\right|_{x=1} = -2 \).

Key Concepts

Derivative at a Point

To understand what a derivative at a point means, let's first recognize that it describes how fast something is changing at a specific location on a curve. Imagine you are standing on a hill. The derivative tells you how steep the hill is at your exact spot. Mathematically, the derivative at a point is expressed as \( \frac{dy}{dx} \) evaluated at a specific \( x \) value. In our exercise, we needed to find the derivative of the function \( y = 1 - x^2 \) at \( x = 1 \). This is like determining the slope of the tangent line to the curve at the specific point \( x = 1 \). By obtaining the derivative equation \( \frac{dy}{dx} = -2x \), we can plug \( x = 1 \) to get \( \left.\frac{dy}{dx}\right|_{x=1} = -2 \). This value \(-2\) indicates that, at \( x = 1 \), the graph of the function is sloping downwards steeply. This idea of evaluating a derivative at a point helps us understand the behavior of functions more deeply.

Basic Rules of Differentiation

Differentiation is the mathematical process of calculating a derivative. The rules of differentiation allow us to find derivatives of functions systematically. Here are some basic rules:

• The Power Rule: If \( y = x^n \), then \( \frac{dy}{dx} = nx^{n-1} \).
• Constant Rule: The derivative of a constant is 0.
• Sum and Difference Rule: The derivative of a sum/difference is the sum/difference of derivatives.

In our exercise, we applied these rules to differentiate \( y = 1 - x^2 \). - First, using the Constant Rule, the derivative of \( 1 \) is \( 0 \). - Then, applying the Power Rule to \( -x^2 \), we get \( \frac{d}{dx}(-x^2) = -2x \). By adding these results, the derivative \( \frac{dy}{dx} = 0 - 2x = -2x \). Understanding these basic rules allows us to quickly and effectively differentiate more complex functions.

Evaluating Derivatives

Evaluating derivatives means plugging a specific \( x \) value into the derivative function to compute the rate of change at that point. This step is crucial because it enables us to quantify the behavior of a function at precise locations. In our problem, after finding the general form of the derivative \( \frac{dy}{dx} = -2x \), we proceeded to evaluate this at \( x = 1 \). Calculating gives us:- Substitute \( x = 1 \).- \( \frac{dy}{dx} = -2 \times 1 = -2 \).This result \(-2\) tells us that at \( x = 1 \), the graph's slope is decreasing sharply. Evaluating derivatives at specific points is essential in physics and engineering for understanding how quantities vary in different scenarios, such as velocity at a moment or stress points in a structure.