To solve for the gradient (slope) of the tangent line to the curve $y = \ln(x)$ that passes through the point $(-2, -1)$, let’s break this into clear steps:

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1. The function and its derivative

The gradient of the tangent line is determined by the derivative of $y = \ln(x)$: $y' = \frac{1}{x}.$

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2. Issue with $\ln(x)$ for negative $x$

The natural logarithm function $\ln(x)$ is only defined for $x > 0$. Since the given point $(-2, -1)$ has $x = -2$, $\ln(x)$ cannot be evaluated at this point. Therefore, there is no tangent line to $y = \ln(x)$ that passes through this point.

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Clarification:

If this was a misinterpretation, and the intention is to find the tangent to another curve (or confirm the problem setup), please clarify.

Would you like to adjust the problem statement, or should I assist further in refining this calculation?

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Related Questions:

1. How is the natural logarithm function defined and restricted?
2. Can tangent lines pass through points outside the domain of the original function?
3. What is the slope of a tangent to $\ln(x)$ at a positive $x$-coordinate?
4. How can we extend $\ln(x)$ to complex numbers and explore tangents?
5. What are the implications of domain and range in logarithmic functions?

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Tip:

Always verify the domain of a function before working with tangent lines or evaluating points!