Problem 1 - Thomas Calculus

Question

In Exercises \(1-12,\) find the first and second derivatives. \begin{equation} y=-x^{2}+3 \end{equation}

Step-by-Step Solution

Answer

The first derivative is \(-2x\), and the second derivative is \(-2\).

Step 1: Identify the Function

The function provided is a simple quadratic equation: \( y = -x^2 + 3 \). Our task is to find the first and second derivatives of this function.

Step 2: Find the First Derivative

To find the first derivative, we differentiate \( y = -x^2 + 3 \) with respect to \( x \). The derivative of \(-x^2\) is \(-2x\), and the derivative of a constant \(3\) is \(0\). Therefore, the first derivative is \( \frac{dy}{dx} = -2x \).

Step 3: Find the Second Derivative

Next, we differentiate the first derivative \( \frac{dy}{dx} = -2x \) to find the second derivative. The derivative of \(-2x\) is \(-2\), as the differentiation of a linear term \(ax\) is simply \(a\). Therefore, the second derivative is \( \frac{d^2y}{dx^2} = -2 \).

Key Concepts

First Derivative

The first derivative of a function reflects how the function's value changes as its input changes, essentially providing the slope of the tangent line to the curve of the function at any given point. For the quadratic function given, \[ y = -x^2 + 3 \]we want to determine how this expression changes with respect to \( x \).- Begin by applying the power rule to differentiate the term \(-x^2\). The power rule states that for any term \( ax^n \), the derivative is \( anx^{n-1} \). Here, \( a = -1 \) and \( n = 2 \), so the derivative becomes \(-2x\).- As for the constant term '3', recall that the derivative of a constant is always zero because constants do not change.Combining these results, the first derivative of the given function is:\[ \frac{dy}{dx} = -2x \]This derivative tells us that the slope of the tangent to the curve is linearly related to the \( x \)-coordinate, and it gets more negative as \( x \) increases. This means the function is concave down.

Second Derivative

The second derivative of a function provides insight into the function's concavity and the behavior of its first derivative. By taking the derivative of the first derivative, we can determine how the rate of change itself is changing.For the first derivative:\[ \frac{dy}{dx} = -2x \]we will differentiate once more:- Since \( -2x \) is a linear function, you can use the power rule again: \(-2x\) becomes \(-2\) because the power of \( x \), which is 1, is reduced to 0, making \( x^0 \) equal to 1.Hence, the second derivative is:\[ \frac{d^2y}{dx^2} = -2 \]This constant negative value indicates that the function is consistently concave down with no points of inflection. In simpler terms, the graph of the original function opens downwards throughout, maintaining a consistent curvature.

Quadratic Function

A quadratic function is a polynomial of degree 2 and can generally be expressed in the standard form:\[ y = ax^2 + bx + c \]In this case, our function is:\[ y = -x^2 + 3 \]which is a special instance where \( a = -1 \), \( b = 0 \), and \( c = 3 \).Some key characteristics of quadratic functions include:- **Graph Shape:** The graph of a quadratic function forms a parabola. If the leading coefficient \( a \) is positive, the parabola opens upwards; if negative, it opens downwards.- **Vertex:** The vertex represents the highest or lowest point of the parabola. For \( y = -x^2 + 3 \), the vertex is at the point \( (0, 3) \), marking the maximum value due to the downward opening of the parabola.Understanding these properties is crucial for analyzing how the function behaves across different values of \( x \), particularly in identifying its intercepts and the direction in which it opens.