# How do you find the equation of a line tangent to the function #y=ln(-x)# at (-2,ln2)?

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To find the equation of a line tangent to the function y=ln(-x) at (-2,ln2), we need to find the slope of the tangent line at that point.

First, we find the derivative of the function y=ln(-x) using the chain rule. The derivative of ln(-x) is -1/x.

Next, we substitute x=-2 into the derivative to find the slope at that point. The slope is -1/(-2) = 1/2.

Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we substitute the values (-2, ln2) and 1/2 into the equation.

Therefore, the equation of the line tangent to the function y=ln(-x) at (-2,ln2) is y - ln2 = (1/2)(x + 2).

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