To solve this, we need to find the change in \( y \) using the given function \( y = x^2 + 7x \), for \( x = 5 \) and \( \Delta x = dx = 0.1 \).

### Step 1: Calculate the original \( y \) at \( x = 5 \)
Substitute \( x = 5 \) into the equation:

\[
y(5) = (5)^2 + 7(5) = 25 + 35 = 60
\]

### Step 2: Use the derivative to estimate \( \Delta y \)
The derivative \( \frac{dy}{dx} \) gives us the rate of change of \( y \) with respect to \( x \). First, we find \( \frac{dy}{dx} \):

\[
\frac{dy}{dx} = \frac{d}{dx}(x^2 + 7x) = 2x + 7
\]

Now, substitute \( x = 5 \) into the derivative to find the rate of change at that point:

\[
\frac{dy}{dx} \bigg|_{x=5} = 2(5) + 7 = 10 + 7 = 17
\]

### Step 3: Estimate \( \Delta y \)
Now use the formula \( \Delta y \approx \frac{dy}{dx} \cdot \Delta x \):

\[
\Delta y \approx 17 \cdot 0.1 = 1.7
\]

So, the change in \( y \) when \( x \) increases by 0.1 is approximately 1.7.

### Step 4: Calculate the new \( y \) value
To estimate the new \( y \) value at \( x = 5.1 \), we add \( \Delta y \) to the original \( y \):

\[
y(5.1) \approx y(5) + \Delta y = 60 + 1.7 = 61.7
\]

Thus, \( y(5.1) \approx 61.7 \).

Would you like more details or have any questions?

Here are some related questions:
1. How does the derivative help estimate small changes in a function?
2. What is the difference between \( \Delta y \) and \( dy \) in calculus?
3. How would the result change if \( \Delta x \) was larger?
4. Can we calculate the exact \( y(5.1) \) without using the derivative approximation?
5. What happens to the rate of change \( \frac{dy}{dx} \) as \( x \) increases further?

**Tip:** Derivatives give us a powerful tool to approximate changes in a function for small values of \( \Delta x \).

Learn how to calculate the change in y using derivatives for the function y = x^2 + 7x at x=5 with a small increment Δx=0.1. This solution explores key calculus concepts such as derivatives and their use in estimating small changes in functions.

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