Differentiate the function y = ln(x).

The derivative of y = ln(x) is 1/x.

To differentiate y = ln(x), we use the formula for the derivative of a natural logarithm function:

d/dx ln(x) = 1/x

This means that the derivative of y = ln(x) is equal to 1/x. To see why this is the case, we can use the definition of the derivative:

f'(x) = lim(h->0) [f(x+h) - f(x)]/h

Applying this to y = ln(x), we get:

y' = lim(h->0) [ln(x+h) - ln(x)]/h

Using the logarithmic identity ln(a) - ln(b) = ln(a/b), we can simplify this expression:

y' = lim(h->0) ln[(x+h)/x]/h

Using the logarithmic identity ln(a/b) = ln(a) - ln(b), we can further simplify:

y' = lim(h->0) [ln(x+h) - ln(x)]/h
= lim(h->0) [ln(x+h)/x - ln(x)/x]/h
= lim(h->0) [ln(1+h/x)]/h - ln(x)/x

Using the limit definition of the natural logarithm ln(1+x) = lim(n->∞) n[(1+x)^(1/n) - 1], we can rewrite the first term as:

y' = lim(h->0) [ln(1+h/x)]/h
= lim(h->0) [1/(h/x)][(1+h/x)^(x/h) - 1]
= lim(h->0) [(1+h/x)^(x/h) - 1]/(h/x)
= lim(h->0) [(1+h/x)^(x/hx) - 1]/(h/x)
= lim(h->0) [(1+h/x)^(1/hx) - 1]/(1/h)
= ln(e) = 1

Therefore, we have:

y' = 1/x

This means that the slope of the tangent line to the graph of y = ln(x) at any point (x, ln(x)) is equal to 1/x.