We must now find the slope of the tangent line at the point of tangency, (1, In 4), by calculating the derivative of s(t) and evaluating the derivative at $t = 1$.
Recall the derivative of ln x.
$\frac{d}{dx}$ ln x =
Use the chain rule to find s'(t).
s(t) = ln(6 - 2t)
s'(t) = $\frac{d}{dt}$
= $\frac{1}{6 - 2t} \cdot \frac{d}{dt}$(
=
Evaluate s'(1).
s'(1) =
