Differential calculus
33
Differentiation of inverse
trigonometric and hyperbolic
functions
33.1 Inverse functions
Ify= 3 x−2, then by transposition,x=
y+ 2
3
. The
functionx=
y+ 2
3
is called theinverse functionof
y= 3 x−2 (see page 201).
Inverse trigonometric functionsare denoted by
prefixing the function with ‘arc’ or, more com-
monly, by using the−^1 notation. For example, if
y=sinx, thenx=arcsin y orx=sin−^1 y. Similarly,
ify=cosx, thenx=arccosyorx=cos−^1 y, and so
on. In this chapter the−^1 notation will be used. A
sketch of each of the inverse trigonometric functions
is shown in Fig. 33.1.
Inverse hyperbolic functionsare denoted by pre-
fixing the function with ‘ar’ or, more commonly, by
using the−^1 notation. For example, ify=sinhx, then
x=arsinhyorx=sinh−^1 y. Similarly, ify=sechx,
thenx=arsechyorx=sech−^1 y, and so on. In this
chapter the−^1 notation will be used. A sketch of
each of the inverse hyperbolic functions is shown in
Fig. 33.2.
33.2 Differentiation of inverse
trigonometric functions
(i) Ify=sin−^1 x, thenx=siny.
Differentiating both sides with respect toy
gives:
dx
dy
=cosy=
√
1 −sin^2 y
since cos^2 y+sin^2 y=1, i.e.
dx
dy
=
√
1 −x^2
However
dy
dx
=
1
dx
dy
Hence, wheny=sin−^1 xthen
dy
dx
=
1
√
1 −x^2
(ii) A sketch of part of the curve ofy=sin−^1 x
is shown in Fig. 33(a). The principal value of
sin−^1 xis defined as the value lying between
−π/2 and π/2. The gradient of the curve
between pointsAandBis positive for all values
ofxand thus only the positive value is taken
when evaluating
1
√
1 −x^2
.
(iii) Given y=sin−^1
x
a
then
x
a
=siny and
x=asiny
Hence
dx
dy
=acosy=a
√
1 −sin^2 y
=a
√[
1 −
(x
a
) 2 ]
=a
√(
a^2 −x^2
a^2
)
=
a
√
a^2 −x^2
a
=
√
a^2 −x^2
Thus
dy
dx
=
1
dx
dy
=
1
√
a^2 −x^2
i.e.wheny=sin−^1
x
a
then
dy
dx
=
1
√
a^2 −x^2
Since integration is the reverse process of
differentiation then:
∫
1
√
a^2 −x^2
dx=sin−^1
x
a
+c
(iv) Giveny=sin−^1 f(x) the function of a function
rule may be used to find
dy
dx
.