Derivation of the Inverse Hyperbolic Trig Functions
[Pages:4]Derivation of the Inverse Hyperbolic Trig Functions
y = sinh-1 x. By definition of an inverse function, we want a function that satisfies the condition
x = sinh y
= ey - e-y by definition of sinh y 2
ey - e-y ey
=
2
ey
=
e2y - 2ey
1
.
2eyx = e2y - 1.
e2y - 2xey - 1 = 0.
(ey)2 - 2x(ey) - 1 = 0.
ey = 2x + 4x2 + 4 2
= x + x2 + 1.
ln(ey) = ln(x + x2 + 1).
y = ln(x + x2 + 1).
Thus
sinh-1 x = ln(x +
Next we compute the derivative of f (x) = sinh-1 x.
x2 + 1).
f (x) =
1
1 + 1 (x2 + 1)-1/2(2x)
x + x2 + 1
2
=
1 . x2 + 1
1
y = cosh-1 x. By definition of an inverse function, we want a function that satisfies the condition
x = cosh y
= ey + e-y by definition of cosh y 2
ey + e-y ey
=
2
ey
=
e2y + 2ey
1
.
2eyx = e2y + 1.
e2y - 2xey + 1 = 0.
(ey)2 - 2x(ey) + 1 = 0.
ey = 2x + 4x2 - 4 2
= x + x2 - 1.
ln(ey) = ln(x + x2 - 1).
y = ln(x + x2 - 1).
Thus
cosh-1 x = ln(x +
Next we compute the derivative of f (x) = cosh-1 x.
x2 - 1).
f (x) =
1
1 + 1 (x2 - 1)-1/2(2x)
x + x2 - 1
2
= 1 . x2 - 1
2
y = tanh-1 x. By definition of an inverse function, we want a function that satisfies the condition
x = tanh y
=
ey - e-y ey + e-y
by definition of
tanh y
ey - e-y ey = ey + e-y ey
=
e2y e2y
- +
1. 1
x(e2y + 1) = e2y - 1.
(x - 1)e2y + (x + 1) = 0.
e2y
=
-
x x
+ -
1 1
.
ln(e2y )
=
ln
-
x x
+ -
1 1
.
2y
=
ln
-
x x
+ -
1 1
.
y
=
1 2
ln
-
x x
+ -
1 1
=
1 2
(ln(x
+
1)
-
ln(-[x
-
1]))
= 1 (ln(x + 1) - ln(1 - x)). 2
Thus
tanh-1 x = 1 (ln(x + 1) - ln(1 - x)). 2
Next we compute the derivative of f (x) = tanh-1 x.
f (x)
=
1 2
x
1 +
1
-
1
1 -
x
(-1)
11
1
= 2 x+1 + 1-x
=
1
1 -
x2
.
3
y = sech-1x. By definition of an inverse function, we want a function that satisfies the condition
x = sechy
=
2 ey + e-y
by definition of sechy
=
2 ey + e-y
ey ey
=
2ey e2y +
1
.
x(e2y + 1) = 2ey.
xe2y - 2ey + x = 0.
ey
=
-(-2) +
(-2)2 - 4(x)(x) 2x
2 + 4(1 - x2)
=
2x
2 + 2 1 - x2
=
2x
=
1+
1 x
- x2
.
y
=
ln
1 + 1 - x2 x
= ln(1 + 1 - x2) - ln x.
Thus
sech-1x = ln(1 +
Next we compute the derivative of f (x) = sech-1x.
1 - x2) - ln x.
f (x)
=
1 1 + 1 - x2
1 (1 - x2)-1/2(-2x) 2
-
1 x
= - 1 . x 1 - x2
4
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