We must now find the slope of the tangent line at the point of tangency, (1, In 2), by aun buneinɔ of s(t) and evaluating the derivative at t = 1. Recall the derivative of In x. In x = dx Use the chain rule to find s'(t). s(t) = In(8 - 6t) s'(t) = dt 1 %3D 8 - 6t %3D Evaluate s'(1). s'(1) = Submit Skip (you cannot come back)

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.6: Quadratic Functions
Problem 50E
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Step 2 of 3
We must now find the slope of the tangent line at the point of tangency, (1, In 2), by calculating the derivative
of s(t) and evaluating the derivative at t = 1.
Recall the derivative of lIn x.
1
d
In x =
dx
Use the chain rule to find s'(t).
s(t) = In(8 – 6t)
s'(t)
dt
8 - 6t
dt
Evaluate s'(1).
s'(1)
Submit
Skip (you cannot come back)
Transcribed Image Text:Step 2 of 3 We must now find the slope of the tangent line at the point of tangency, (1, In 2), by calculating the derivative of s(t) and evaluating the derivative at t = 1. Recall the derivative of lIn x. 1 d In x = dx Use the chain rule to find s'(t). s(t) = In(8 – 6t) s'(t) dt 8 - 6t dt Evaluate s'(1). s'(1) Submit Skip (you cannot come back)
Tutorial Exercise
Find an equation of the tangent line at the point indicate.
s(t) = In(8 6t),
t = 1.
Step 1 of 3
Recall that a tangent line to a graph of y = f(x) at a point P(a, f(a)) is the line through the point P of slope f
f(a) =
(a
f '(a)
The equation of the tangent line in point-slope form is y - f(a)
f '(a) (x - a).
To find the equation of the tangent line to the graph of s(t) = In(8 6t), we will need to find the slope of the
tangent line to the graph of s(t) at the point (1, s(1)).
Find s(1).
s(1) = In(8 – 6(1)) = |In(2)
In(2)
(1, In(2)
1, In(2)
Therefore, the point of tangency is (t, s(t)) = (1, s(1)) =
Transcribed Image Text:Tutorial Exercise Find an equation of the tangent line at the point indicate. s(t) = In(8 6t), t = 1. Step 1 of 3 Recall that a tangent line to a graph of y = f(x) at a point P(a, f(a)) is the line through the point P of slope f f(a) = (a f '(a) The equation of the tangent line in point-slope form is y - f(a) f '(a) (x - a). To find the equation of the tangent line to the graph of s(t) = In(8 6t), we will need to find the slope of the tangent line to the graph of s(t) at the point (1, s(1)). Find s(1). s(1) = In(8 – 6(1)) = |In(2) In(2) (1, In(2) 1, In(2) Therefore, the point of tangency is (t, s(t)) = (1, s(1)) =
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