Section 3.9: Inverse Trig Derivatives 1 Derivative of sin (x)

Section 3.9: Inverse Trig Derivatives
1
Derivative of sin−1 (x)
Let’s examine the equation
y = sin−1 (x)
and find y 0 . We can solve this equation for x, giving
sin(y) = x
Differentiating on both sides gives
cos(y)y 0 = 1
or
y0 =
1
cos(y)
We now use the trig identity sin2 (y) + cos2 (y) = 1, which gives that
q
cos(y) = 1 − sin2 (y)
But, from the original equation, we know that sin(y) = x, so we have that
p
cos(y) = 1 − x2
so
y0 = √
2
1
1 − x2
Others
Using this method on the other trig inverse functions we obtain the following
table of deriatives:
y0
1
√
sin−1 (x)
1 − x2
1
cos−1 (x)
−√
1 − x2
1
tan−1 (x)
1 + x2
1
cot−1 (x)
−
1 + x2
1
−1
√
sec (x)
|x| x2 − 1
1
csc−1 (x) − √
|x| x2 − 1
y
1
3
General Inverses
Suppose we have a point (x0 , y0 ), know f is invertible (f −1 exists) and we know
f 0 (x0 ). Then we have the following relation:
(f −1 )0 (y0 ) =
4
1
f 0 (x0 )
Homework
1. Differentiate y = sin−1 (5x4 )
Solution: Using the chain rule: y 0 =
1
(20x3 )
1 − (5x4 )2
2. Differentiate y = cos(sin−1 (7x))
−1
Solution: Using Chain Rule: y 0 = − sin(sin
(7x))
1
p
1 − (7x)2
!
(7)
which equals
y0 = − √
49x
1 − 49x2
since sin and sin−1 are inverses of each other.
3. A boat sails directly toward a 150-m skyscraper that stands on the edge
of a harbor. The angular size of the building is the angle formed by the
dθ
lines from the top and bottom of the building to the observer. What
dx
when the boat is x = 500m from the building?
Solution θ = tan−1
150
, so, using the chain rule,
x
dθ
1
150
=
2 − 2
dx
x
150
1+
x
Plug in x = 500
1
2
150
1+
500
2
−
150
5002