Calculus Of One Real Variable –
By Pheng Kim Ving 
7.7 
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1. Definitions 
The graph of the hyperbolic sine
function y = sinh x is sketched in Fig. 1.1. Clearly sinh
is onetoone, and so has an
inverse, denoted sinh^{–1}. The inverse hyperbolic sine
function sinh^{–1} is defined as follows:
The graph of y = sinh^{–1} x is the mirror image of that of y = sinh x
in the line y = x.
It's shown in Fig. 1.1. We have
dom(sinh^{–1})
= R and range(sinh^{–1}) = R.
Fig. 1.1 Graph of y = sinh^{–1} x. 
Fig. 1.2 Graph of y = cosh^{–1} x. 
The graph of the hyperbolic tangent function y = tanh x
is sketched in Fig. 1.3. Clearly tanh is onetoone, and so
has an
inverse, denoted tanh^{–1}. The inverse hyperbolic tangent
function tanh^{–1} is defined as follows:
Fig. 1.3 Graph of y = tanh^{–1} x. 
The graph of the hyperbolic cotangent function y = coth x
is sketched in Fig. 1.4. Clearly coth is onetoone, and
thus has
an inverse, denoted coth^{–1}. The inverse hyperbolic cotangent
function coth^{–1} is defined as follows:
Fig. 1.4 Graph of y = coth^{–1} x. 
Fig. 1.5 Graph of y = sech^{–1} x. 
The graph of the hyperbolic cosecant function y = csch x
is sketched in Fig. 1.6. Clearly csch is onetoone, and so
has
an inverse, denoted csch^{–1}. The inverse hyperbolic cosecant
function csch^{–1} is defined as follows:
Fig. 1.6 Graph of y = csch^{–1} x. 
Prove the identity:
Recall that the inverse of the natural exponential function
is the natural logarithm function. Since the hyperbolic functions
are defined in terms of the natural exponential function, it's not surprising
that their inverses can be expressed in terms
of the natural logarithm function. Also see Problem
& Solution 1 and Problem & Solution 2.
Let y = sinh^{–1} x. Then x = sinh
y = (e^{y}
– e^{–}^{y})/2.
So e^{y} – e^{–}^{y} – 2x
= 0. Multiplying both sides by e^{y}
yields e^{2}^{y} – 1 –
2xe^{y} = 0,
or e^{2}^{y} – 2xe^{y} – 1 = 0, which is a quadratic
equation in e^{y}. Its roots are:
EOS
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2. Differentiation 

We prove formula [2.1] as follows. Let y
= sinh^{–1}
x. Then x = sinh
y. Differentiating this equation implicitly with
respect to x
we get:
The remaining differentiation formulas are proved in a similar way.
Differentiate sinh^{–1} tan x.
Problems & Solutions 
1. In Example 1.1 we proved the identity:
Also see Problem & Solution 2.
Solution
Let y = cosh^{–1} x. Then x = cosh
y = (e^{y}
+ e^{–}^{y})/2. So e^{y} + e^{–}^{y}
– 2x = 0. Multiplying both sides by e^{y} yields e^{2}^{y} + 1 – 2xe^{y} =
0, or e^{2}^{y} – 2xe^{y} + 1 = 0, which is a
quadratic equation in e^{y}.
Its roots are:
2. In Example 1.1 we proved 1 identity and in Problem & Solution 1 you were asked to prove
another identity. Now again
you're asked to prove the following
2 identities:
Solution
a. Let y = tanh^{–1} x. So x = tanh y and x < 1. We have:
xe^{2}^{y} – x = e^{2}^{y} + 1,
e^{2}^{y}(x – 1) = x + 1,
3. Differentiate the following functions.
a. sinh^{–1} (x/a), a > 0.
b. cosh^{–1} (x/a), a > 0.
Solution
4. Differentiate the following functions.
a. y = sech^{–1} (x^{2}).
b. f(t) = csch^{–1} tan t.
Solution
5. Prove that:
Solution
Let y = csch^{–1} x. Then x = csch y. Let z = sinh^{–1} (1/x), so that 1/x = sinh z, or:
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