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Calculus 1 Problems & Solutions – Chapter 4 – Section 4.3.2 |
4.3.2
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Review |
1. Definitions |
The graph of the hyperbolic sine function y = sinh x is sketched
in Fig. 1.1. Clearly sinh
is one-to-one, 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.
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Fig. 1.1
Graph of y = sinh–1 x. |
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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 one-to-one, and so has an
inverse, denoted tanh–1. The inverse hyperbolic tangent
function tanh–1 is defined as follows:
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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 one-to-one, and thus has
an inverse, denoted coth–1. The inverse hyperbolic cotangent
function coth–1 is defined as follows:
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Fig. 1.4
Graph of y = coth–1 x. |
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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 one-to-one, and so has
an inverse, denoted csch–1. The inverse hyperbolic cosecant function
csch–1 is defined as follows:
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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 = (ey
– e–y)/2. So ey
– e–y – 2x = 0. Multiplying both sides
by ey
yields e2y – 1 – 2xey
= 0,
or e2y – 2xey – 1 = 0, which
is a quadratic equation in ey.
Its roots are:
EOS
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2. Differentiation |
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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 = (ey + e–y)/2.
So ey + e–y
– 2x
= 0. Multiplying both sides by ey
yields e2y
+ 1 – 2xey =
0, or e2y
– 2xey + 1 = 0, which is a
quadratic equation in ey. 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:
xe2y – x = e2y + 1,
e2y(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 (x2).
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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