Calculus I, Section 3.4, #26 The Chain Rule Find the derivative of

Calculus I, Section 3.4, #26
The Chain Rule
Find the derivative of the function.1
s
1 + sin (t)
s(t) =
1 + cos (t)
s(t) =
s
1 + sin (t)
=
1 + cos (t)
1 + sin (t)
1 + cos (t)
1 + sin (t)
1 + cos (t)
1/2−1
1 + sin (t)
1 + cos (t)
−1/2
1/2
outside function
inside function
1
ds
=
dt
2
=
1
2
=
1 (1 + sin (t))−1/2 cos (t) + sin (t) + 1
2
2 (1 + cos (t))−1/2
(1 + cos (t))
(1 + cos (t)) · cos (t) − (1 + sin (t)) · − cos (t)
·
2
(1 + cos (t))
cos (t) + cos2 (t) + sin (t) + sin2 (t)
1/2
cos (t) + sin (t) + 1
1 (1 + cos (t))
1/2
2 (1 + sin (t))
(1 + cos (t))2
cos (t) + sin (t) + 1
= p
3/2
2 1 + sin (t) (1 + cos (t))
=
1 Stewart,
2
(1 + cos (t))
Calculus, Early Transcendentals, p. 204, #26.