Problem 33 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The derivative \( dy/dx \) at \( x = 1 \) is \( -2 \).

Step 1: Understand the Function

We are given the function \( y = 1 - x^2 \). This is a quadratic function, and our task is to find its derivative with respect to \( x \), and then evaluate it at \( x = 1 \).

Step 2: Differentiate the Function

To find the derivative of \( y = 1 - x^2 \) with respect to \( x \), use the power rule. The derivative of \( x^n \) is \( n \cdot x^{n-1} \). Here, \( n \) is 2 for \( x^2 \). Therefore, \( \frac{d}{dx}(x^2) = 2x \). The constant \( 1 \) becomes zero. So the derivative is \( \frac{dy}{dx} = -2x \).

Step 3: Evaluate the Derivative at x=1

Substitute \( x = 1 \) into the derivative \( \frac{dy}{dx} = -2x \). This means substituting 1 for \( x \), giving us \( -2 \times 1 = -2 \).

Key Concepts

Quadratic Function

The term "quadratic function" may sound complex, but it's really just a polynomial where the highest power of the variable is two. The general form of a quadratic function is:

• \( y = ax^2 + bx + c \)

In our exercise, the function is given as \( y = 1 - x^2 \). That means:

• The coefficient \( a \) of \( x^2 \) is \(-1\).
• The coefficient \( b \) of \( x \) is \( 0 \) because there is no \( x \) term.
• The constant term \( c \) is \( 1 \).

Quadratic functions graph into a parabola. In this case, it's an upside-down opening parabola because the coefficient \( a \) is negative. Quadratic functions are important in calculus because they represent curved lines, which are essential in understanding changes and behaviors of graphs. Differentiating these can tell us how steep the curve is or how it changes over time.

Power Rule

The power rule is a fundamental tool in calculus that helps us differentiate functions involving powers of a variable. The rule states that if you have a function \( x^n \), its derivative with respect to \( x \) is:

• \( n \cdot x^{n-1} \)

This means you multiply the term by its exponent and then decrease the exponent by one.
In our exercise, we're working with \( y = 1 - x^2 \), so to find the derivative, we apply the power rule to \( x^2 \). Here:

• \( n \) is 2, so the derivative is \( 2x \).

But remember, since our original function has a negative sign in front, the derivative actually becomes \(-2x \).
Using the power rule speeds up the process of differentiation and allows us to handle polynomial functions efficiently.

Evaluating Derivatives

Once we've differentiated a function, evaluating that derivative at a specific point gives us valuable information. It tells us the rate of change or the slope of the tangent line at that specific point on the curve.
In our problem, after differentiating \( y = 1 - x^2 \) to get \( -2x \), we're asked to evaluate the derivative at \( x = 1 \). This involves replacing \( x \) in \( -2x \) with 1:

• \( -2 \times 1 = -2 \)

This result means that at \( x = 1 \), the slope of the tangent line to the curve \( y = 1 - x^2 \) is \(-2\).
In simpler terms, if you look at the curve at \( x = 1 \), it is decreasing, as indicated by the negative slope value. Evaluating derivatives can help in understanding how quickly a graph is rising or falling at any given point.