Derivative Practice: Inverse Trigonometric Functions

Derivative Practice: Inverse Trigonometric Functions

1. If k(t) = 2arcsin( t), then what is k (t)?

Solution:

k (t) = (ln 2)2arcsin( t) ?

1

1

- ( t)2

?

1

t-

1 2

2

=

ln 2

? 2arcsin( t)

?

1

2

t 1-t

p2 2. If g(p) = arctan(5p - 1) + k, then what is g (p)?

3

Solution:

p2

1

2p

g (p) = 3 1 + (5p - 1)2 ? 5 + 3 arctan(5p - 1)

x 3. If f (x) = arcsin (ex) , then what is f (x)?

Solution:

f

(x)

=

arcsin

(ex)

-

x

?

ex 1-e2x

[arcsin (ex)]2

4. If h(x) = tan(arctan(x)), then what is h (x)?

Solution:

sec2(arctan(x))

h (x) =

1 + x2

Better solution: We have h(x) = tan(arctan(x)) = x by inverse functions. So, h (x) = 1.

Note that by using the triangle technique, the first solution can be simplified:

1 + x2

x

= arctan(x)

hyp 1 + x2 sec() = =

1

adj

1

sec2(arctan(x)) = (sec())2 = 1 + x2

sec2(arctan(x)) 1 + x2

1 + x2

= 1 + x2 = 1

Happily, the two methods of finding the derivative yield the same answer.

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