Use the formula for instantaneous rate of​ change, approximating the limit by using smaller and smaller values of​ h, to find the instantaneous rate of change for the given function at the given value.
f left parenthesis x right parenthesis equals 2 x Superscript x​;  xequals2
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Part 1
The instantaneous rate of change for the function at xequals2 is 

To find the instantaneous rate of change of the function \( f(x) = 2^x \) at \( x = 2 \), we use the limit definition of the derivative, which is:

\[
f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}
\]

For \( f(x) = 2^x \) at \( x = 2 \), this becomes:

\[
f'(2) = \lim_{h \to 0} \frac{2^{2 + h} - 2^2}{h}
\]

Now, let's approximate this limit by using smaller and smaller values of \( h \). I'll calculate it for \( h = 0.1, 0.01, \) and \( 0.001 \) to get progressively closer to the actual rate of change.

We know \( f(2) = 2^2 = 4 \).

Now let's compute \( f(2 + h) \) for different values of \( h \):

1. **For \( h = 0.1 \):**
   \[
   f(2.1) = 2^{2.1} \approx 4.287
   \]
   \[
   \frac{4.287 - 4}{0.1} \approx 2.87
   \]

2. **For \( h = 0.01 \):**
   \[
   f(2.01) = 2^{2.01} \approx 4.028
   \]
   \[
   \frac{4.028 - 4}{0.01} \approx 2.82
   \]

3. **For \( h = 0.001 \):**
   \[
   f(2.001) = 2^{2.001} \approx 4.0028
   \]
   \[
   \frac{4.0028 - 4}{0.001} \approx 2.82
   \]

From these approximations, we see that the instantaneous rate of change approaches \( 2.8 \) as \( h \) becomes smaller. Hence, the instantaneous rate of change for \( f(x) = 2^x \) at \( x = 2 \) is approximately:

\[
f'(2) \approx 2.8
\]

### Would you like further details or have any questions?

Here are five related questions for you to consider:
1. How does the derivative of exponential functions differ from polynomial functions?
2. Can you explain why we use smaller and smaller values of \( h \) for the approximation?
3. What is the significance of the derivative in real-world applications?
4. How would the derivative change if we used a different base, such as \( 3^x \)?
5. What does the instantaneous rate of change represent geometrically?

**Tip:** The derivative gives you the slope of the tangent line at any given point on the function graph.

Use the formula for instantaneous rate of change, approximating the limit by using smaller and smaller values of h, to find the instantaneous rate of change for the given function at the given value. f(x) = 2^x; x = 2.

Learn how to find the instantaneous rate of change of the exponential function f(x) = 2^x at x = 2 using the limit definition of the derivative. This detailed solution covers step-by-step calculations, approximating limits with smaller values of h. Suitable for high school and early college calculus students.

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