Question

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) =

          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) =

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.
(d)/(dx) ln x =
Use the chain rule to find s'(t).
s(t) = ln(6 - 2t)
s'(t) = (d)/(dt)
= (1)/(6 - 2t)·(d)/(dt)(
=
Evaluate s'(1).
s'(1) =

Added by James J.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Andrew Noble

03/26/2023

Step 1

Step 1: Apply the chain rule to find the derivative of s(t): s'(t) = d/dt(ln(6-2t)) Using the chain rule, we have: s'(t) = (1/(6-2t)) * (-2) s'(t) = -2/(6-2t) Show more…

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We must now find the slope of the tangent line at the point of tangency, (1, ln4), by calculating the derivative of s(t) and evaluating the derivative at t=1. Recall the derivative of ln(x): (d/dx)ln(x) = 1/x Use the chain rule to find s'(t): s'(t) = d/dt(ln(6-2t)) Evaluate s'(1): s'(1) = 1/(6-2(1)) = 1/4