To estimate the graph of $f(x) = 2^x$, we recognize that it's an exponential function. The graph is continuous and increasing because $2^x$ grows as $x$ increases. It passes through $(0,1)$, since $2^0 = 1$, and increases rapidly for positive $x$. For negative values of $x$, the function approaches zero but never reaches it (asymptote at $y = 0$).

Now, let’s focus on finding a formula for $f'(x)$.

Derivative of $f(x) = 2^x$

You're given the difference quotient approximation for the derivative of $f(x)$ at $x = a$: $f'(a) \approx \frac{2^{a + 0.000001} - 2^{a - 0.000001}}{0.000002}$

This form suggests the standard definition of a derivative: $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x - h)}{2h}$ However, let's derive the exact formula for the derivative of $2^x$ using the rules of exponential functions.

1. Start with the exponential function: $f(x) = 2^x$

2. Recall that the derivative of an exponential function of the form $a^x$ is: $\frac{d}{dx} a^x = a^x \ln(a)$ where $a$ is a constant (in this case, $a = 2$).

3. Applying this rule to $2^x$: $f'(x) = 2^x \ln(2)$

So, the derivative of $f(x) = 2^x$ is: $f'(x) = 2^x \ln(2)$

Summary:

The derivative of $f(x) = 2^x$ is: $f'(x) = 2^x \ln(2)$

This means that the slope of the function at any point $x$ is proportional to the value of $2^x$, scaled by the constant $\ln(2)$, which is approximately 0.6931.

Would you like more details or explanations on any part of this?

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Here are five related questions to expand on this topic:

1. How would the derivative change if $f(x) = 3^x$ instead of $2^x$?
2. What is the second derivative of $f(x) = 2^x$?
3. How does the growth rate of $2^x$ compare to $e^x$?
4. Can you explain why $\ln(2)$ appears in the derivative of $2^x$?
5. How would you estimate the tangent line at a specific point on the graph of $2^x$?

Tip: When dealing with exponential functions, remember that the derivative involves the natural logarithm of the base.