DIFFERENTIATION OF INVERSE TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS 337G
- (a)θ^2 cos−^1 (θ^2 −1) (b) (1−x^2 ) tan−^1 x
⎡
⎢
⎢
⎣(a) 2θcos−^1 (θ^2 −1)−2 θ^2
√
2 −θ^2(b)(
1 −x^2
1 +x^2)
− 2 xtan−^1 x⎤⎥
⎥
⎦- (a) 2
√
tcot−^1 t (b)xcosec−^1√ x ⎡ ⎢ ⎢ ⎣(a)− 2√
t
1 +t^2+1
√
tcot−^1 t(b) cosec−^1√
x−1
2√
(x−1)⎤⎥
⎥
⎦- (a)
sin−^13 x
x^2(b)cos−^1 x
√
1 −x^2
⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
(a)1
x^3{
3 x
√
1 − 9 x^2−2 sin−^13 x}(b)− 1 +x
√
1 −x^2cos−^1 x(1−x^2 )⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦33.3 Logarithmic forms of the inverse
hyperbolic functions
Inverse hyperbolic functions may be evaluated most
conveniently when expressed in a logarithmic
form.
For example, ify=sinh−^1
x
athenx
a=sinhy.From Chapter 5, ey=coshy+sinhyand
cosh^2 y−sinh^2 y=1, from which,
coshy=
√
1 +sinh^2 ywhich is positive since coshy
is always positive (see Fig. 5.2, page 43).
Hence ey=√
1 +sinh^2 y+sinhy=√[1 +(xa) 2 ]
+x
a=√(
a^2 +x^2
a^2)
+x
a=√
a^2 +x^2
a+x
aorx+√
a^2 +x^2
aTaking Napierian logarithms of both sides gives:
y=ln{
x+√
a^2 +x^2
a}Hence, sinh−^1x
a=ln{
x+√
a^2 +x^2
a}(1)Thus to evaluate sinh−^13
4, letx=3 anda=4in
equation (1).Then sin h−^13
4=ln{
3 +√
42 + 32
4}=ln(
3 + 5
4)
=ln 2= 0. 6931By similar reasoning to the above it may be
shown that:cosh−^1x
a=ln{
x+√
x^2 −a^2
a}and tanh−^1x
a=1
2ln(
a+x
a−x)Problem 9. Evaluate, correct to 4 decimal
places, sinh−^1 2.From above, sinh−^1x
a=ln{
x+√
a^2 +x^2
a}Withx=2 anda=1,sinh−^12 =ln{
2 +√
12 + 22
1}=ln (2+√
5)=ln 4. 2361= 1. 4436 ,correct to 4 decimal placesProblem 10. Show that
tanh−^1x
a=1
2ln(
a+x
a−x)
and evaluate, correctto 4 decimal places, tanh−^13
5Ify=tanh−^1x
athenx
a=tanhy.From Chapter 5,tanhy=sinhx
coshx=1
2 (ey−e−y)
1
2 (ey+e−y)=e^2 y− 1
e^2 y+ 1by dividing each term by e−y