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)
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
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.3: The Natural Exponential Function
Problem 44E
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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)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb34d396a-e853-4145-9f5e-1843400f5055%2F8beb11ec-8711-4093-8f76-218b5e9bdfa0%2F2vk8me_processed.jpeg&w=3840&q=75)
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)) =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb34d396a-e853-4145-9f5e-1843400f5055%2F8beb11ec-8711-4093-8f76-218b5e9bdfa0%2F8e7nrm8_processed.jpeg&w=3840&q=75)
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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