To determine the instantaneous rate of change of $f(x) = 2\ln(x + 5)$ at $x = 2$, we use the limit definition of the derivative:

$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}.$

Here, $a = 2$. Plugging in the given function, we have:

$f'(2) = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h}.$

Step 1: Substitute $f(x)$ into the limit

The function $f(x) = 2\ln(x + 5)$. Substitute this into the limit:

$f'(2) = \lim_{h \to 0} \frac{2\ln((2 + h) + 5) - 2\ln(2 + 5)}{h}.$

Simplify the arguments of the logarithms:

$f'(2) = \lim_{h \to 0} \frac{2\ln(7 + h) - 2\ln(7)}{h}.$

Step 2: Factor out the constant 2

Factor out the constant $2$ from the numerator:

$f'(2) = 2 \cdot \lim_{h \to 0} \frac{\ln(7 + h) - \ln(7)}{h}.$

Step 3: Apply the logarithmic difference rule

Using the property of logarithms $\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$, rewrite the numerator:

$f'(2) = 2 \cdot \lim_{h \to 0} \frac{\ln\left(\frac{7 + h}{7}\right)}{h}.$

At this point, the expression is ready for evaluation using advanced techniques (e.g., L'Hôpital's Rule or series expansion), but since the problem asks not to simplify, this is the final result:

$f'(2) = 2 \cdot \lim_{h \to 0} \frac{\ln\left(\frac{7 + h}{7}\right)}{h}.$

Let me know if you'd like to proceed with simplification or any clarifications!

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Related Questions

1. How would you simplify the limit to find the exact derivative?
2. What is the general derivative of $f(x) = 2\ln(x + 5)$?
3. How does the chain rule apply to differentiating logarithmic functions?
4. What is the instantaneous rate of change geometrically for this function?
5. How would you approximate the rate of change numerically at $x = 2$?

Tip

When using the limit definition of the derivative, always substitute the function directly and carefully handle algebraic simplifications to avoid errors.