Derivative Practice: Inverse Trigonometric Functions
√
1. If k(t) = 2arcsin( t) , then what is k 0 (t)?
Solution:
√
1
1 1
k 0 (t) = (ln 2)2arcsin( t) · q
· t− 2
√
2
1 − ( t)2
=
ln 2 arcsin(√t)
1
·√√
·2
2
t 1−t
p2
arctan(5p − 1) + k, then what is g 0 (p)?
3
Solution:
1
2p
p2
·5+
arctan(5p − 1)
g 0 (p) =
3 1 + (5p − 1)2
3
2. If g(p) =
x
, then what is f 0 (x)?
arcsin (ex )
Solution:
ex
arcsin (ex ) − x · √1−e
2x
f 0 (x) =
x
2
[arcsin (e )]
3. If f (x) =
4. If h(x) = tan(arctan(x)), then what is h0 (x)?
Solution:
h0 (x) =
sec2 (arctan(x))
1 + x2
Better solution: We have h(x) = tan(arctan(x)) = x by inverse functions. So, h0 (x) = 1.
Note that by using the triangle technique, the first solution can be simplified:
θ = arctan(x)
2
√ 1+x
θ
1
x
hyp
sec(θ) =
=
adj
√
1 + x2
1
sec2 (arctan(x)) = (sec(θ))2 = 1 + x2
sec2 (arctan(x))
1 + x2
=
=1
2
1+x
1 + x2
Happily, the two methods of finding the derivative yield the same answer.