What is the slope of the line tangent to the graph of y = ln(1 + x) at x = 1?

To find the slope of the line tangent to the curve y = ln(1 + x) at the point where x = 1, we first need to determine the derivative of the function.

The derivative of y = ln(1 + x) with respect to x can be calculated using the chain rule. The derivative is:

y’ = (1 / (1 + x)) * (d/dx(1 + x))

Since the derivative of 1 + x is simply 1, we have:

y’ = 1 / (1 + x)

Next, we substitute x = 1 into the derivative to find the slope of the tangent line at this specific point:

y'(1) = 1 / (1 + 1) = 1 / 2

Therefore, the slope of the line tangent to the graph of y = ln(1 + x) at x = 1 is 1/2.

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