Problem 17 - Calculus Early Transcendentals

Question

Find \(y^{\prime}(1)\) $$ y=5 x^{2}-3 x+1 $$

Step-by-Step Solution

Answer

The value of \(y'(1)\) is 7.

Step 1: Understand the problem

We need to find the derivative of the function \(y = 5x^2 - 3x + 1\), and then evaluate this derivative at \(x = 1\).

Step 2: Differentiate the function

To find the derivative \(y'\) of the function \(y = 5x^2 - 3x + 1\), apply the power rule, which states that \(\frac{d}{dx} [x^n] = nx^{n-1}\). The derivative of \(5x^2\) is \(10x\), the derivative of \(-3x\) is \(-3\), and the derivative of the constant \(1\) is \(0\). Thus, we find \(y' = 10x - 3\).

Step 3: Evaluate the derivative at x=1

Now that we have the derivative \(y' = 10x - 3\), substitute \(x = 1\) to find \(y'(1)\). This gives us \(y'(1) = 10(1) - 3 = 10 - 3 = 7\).

Key Concepts

Power Rule

The power rule is one of the most fundamental concepts in calculus, especially when working with derivatives. This rule provides a straightforward way to differentiate expressions that involve powers of a variable. If you have a term in the form of \(x^n\), applying the power rule means that you bring the exponent \(n\) down as a coefficient and then subtract 1 from the exponent, resulting in \(nx^{n-1}\).

Consider the term \(5x^2\) in the given function. Using the power rule, you multiply the coefficient (5) by the power (2), giving you \(5 \times 2 = 10\), and then reduce the power by one, resulting in the term \(10x\).

The power rule can also be applied to linear terms like \(-3x\), where the exponent is implicitly 1. Differentiating \(-3x\) gives \(-3\) because the process results in \(-3 \times 1 = -3\) and \(x^{1-1} = x^0 = 1\). It's important to remember that the derivative of a constant, like 1, is zero since constants do not change, thus their rate of change is zero.

Differentiation

Differentiation is the process of finding the derivative of a function. It measures how a function changes as its inputs change. In simple words, it helps you understand the rate of change or the slope of the curve represented by a function.

The function we are examining is \(y = 5x^2 - 3x + 1\). Through differentiation, we aim to determine \(y'\), the derivative of \(y\). By differentiating each term separately:

• For \(5x^2\), using the power rule gives \(10x\).
• For \(-3x\), differentiation provides \(-3\).
• The constant 1, when differentiated, becomes 0 since it doesn’t change.

Combining these results provides the overall derivative \(y' = 10x - 3\). Differentiation becomes especially powerful when interpreting real-life situations because it shows the trend or inclination at any given point. When learning calculus, grasping the concept of differentiation early is key to understanding more complex concepts later on.

Evaluating Derivatives

Once the derivative \(y'\) is determined, evaluating it at a specific point helps find the rate of change at that particular place. This evaluation process is used to determine how fast or slow a function is changing at any specific moment.

In this exercise, we aim to evaluate \(y'\) at \(x = 1\). With the derivative already found as \(y' = 10x - 3\), substitute \(x = 1\) into the expression. Doing this calculation, we obtain: \[y'(1) = 10(1) - 3 = 10 - 3 = 7.\]

This result, 7, tells us that at the point \(x = 1\), the rate of change or slope of the tangent to the curve \(y = 5x^2 - 3x + 1\) is 7. Evaluating derivatives like this is crucial in calculus as it gives precise information about the behavior of a function at a specific point, such as the steepness and direction of the curve.