Problem 17 - Calculus - AP Edition

Question

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

Step-by-Step Solution

Answer

\( y'(1) = 7 \)

Step 1: Identify the Function

We need to find the derivative of the given function. The function is \( y = 5x^2 - 3x + 1 \).

Step 2: Differentiate the Function

Differentiate the function \( y = 5x^2 - 3x + 1 \) with respect to \( x \) to find \( y' \). Use the power rule: \( \frac{d}{dx}[x^n] = nx^{n-1} \).Differentiate term by term:- The derivative of \( 5x^2 \) is \( 10x \).- The derivative of \( -3x \) is \( -3 \).- The derivative of constant \( 1 \) is \( 0 \).Thus, \( y' = 10x - 3 \).

Step 3: Evaluate the Derivative at \( x=1 \)

Substitute \( x = 1 \) into the derivative \( y' = 10x - 3 \).\( y'(1) = 10(1) - 3 = 10 - 3 = 7 \).

Step 4: Conclusion

The value of the derivative at \( x = 1 \) is \( 7 \).

Key Concepts

Power Rule

When you're dealing with polynomials in calculus, the power rule is your best friend for differentiation. It's a super handy technique because it makes finding derivatives nearly automatic. The power rule states that if you have a function of the form \( f(x) = x^n \), its derivative \( f'(x) \), is given by \( nx^{n-1} \).

This rule simplifies the process of differentiation because it lets you "bring down" the exponent as a coefficient and "decrease" the exponent by one. Let's see it in action with an example:

• Given \( f(x) = 5x^2 \), the derivative \( f'(x) \) is \( 2 \times 5x^{2-1} = 10x \).
• For a function like \( g(x) = -3x \), the exponent of \( x \) is one. Applying the power rule here gives us \( -3 \times 1x^{1-1} = -3 \).
• Lastly, when differentiating a constant term like 1, remember the derivative is zero because constants don't change.

This rule is incredibly efficient for polynomials, saving you from more complex differentiation methods. It's also essential for further calculus studies.

Finding the Derivative

Finding the derivative of a function essentially means you're determining the rate at which that function changes. It's like finding the function's slope at any point. For the function \( y = 5x^2 - 3x + 1 \), your job is to differentiate it in terms of \( x \).

Let's break down how we find the derivative of each part using the power rule:

• From \( 5x^2 \), we use the power rule to get \( 10x \).
• From \( -3x \), again use the power rule, and we get \( -3 \).
• The derivative of the constant \( 1 \) is \( 0 \).

So putting it all together, the derivative \( y' \) is \( 10x - 3 \). This equation tells us the slope or rate of change of the function \( y \) at any given value of \( x \). Differentiation can seem like magic, but it's a straightforward and systematic process.

Evaluating Derivative at a Point

Once you have the derivative expression, the next step is often to evaluate it at a specific point. This is especially useful for understanding things like instantaneous rate of change or slope at a particular value of \( x \).

In our exercise, we want to evaluate \( y' = 10x - 3 \) at \( x = 1 \). This means you plug \( x = 1 \) into the derivative:

• First, substitute \( 1 \) for \( x \) in the derivative: \( y'(1) = 10(1) - 3 \).
• Simplify the expression: \( 10(1) = 10 \); therefore, \( y'(1) = 10 - 3 \).
• Finally, \( y'(1) = 7 \). This means the slope of the tangent line to the function at \( x = 1 \) is 7.

This step involves straightforward substitution and arithmetic. Evaluating derivatives at specific points is critical since it gives you precise values of slopes or rates, which is very useful in real-world applications like physics or economics.