MATH 155 SUPPLEMENTAL NOTES 10

INVERSE HYPERBOLIC FUNCTIONS
HYPERBOLIC FUNCTIONS AND THEIR INVERSES

HYPERBOLIC FUNCTIONS	INVERSE HYPERBOLIC FUNCTIONS
y = sinh x	y = sinh ^{- 1}x
y = cosh x (x ³ 0)	y = cosh ^{- 1} x
y = sech x (x ³ 0)	y = sech ^{- 1} x
y = tanh x	y = tanh ^{- 1} x
y = coth x	y = coth ^{- 1} x
y = csch x	y = csch ^{- 1} x

DERIVATIVES OF THE INVERSE HYPERBOLIC FUNCTIONS

EXAMPLE 1:

SOLUTION:

EXAMPLE 2:

SOLUTION:

EXAMPLE 3:

SOLUTION:

EXAMPLE 4:

SOLUTION:

EXAMPLE 5:

SOLUTION:

EXAMPLE 6:

SOLUTION:

Notice that to find the derivatives of the inverse hyperbolic functions requires the knowledge of the formulas for these derivatives. Most of these derivatives will also involve the chain rule. I know that this looks like plug-in-chug, but working with these types of derivatives, this what you will expect. There is a lot of pattern recognition in mathematics and calculus, and this is the case when finding the derivatives of the inverse trig and inverse hyperbolic trig functions.

INTEGRALS RELATED TO THE INVERSE HYPERBOLIC TRIG FUNCTIONS

Which formula do you use? Look at the integral and notice the pattern. The pattern will help you determine the formula that you need to use.

EXAMPLE 7:

SOLUTION:

Here u = x and a = 3, and x ² + 9 = u ² + a ². Therefore, the only pattern that fits this integral is the inverse hyperbolic sine function.

EXAMPLE 8:

SOLUTION:

Let a = 3 and u = 2x, that implies du = 2dx. 4x ² - 9 = u ² - a ², therefore the pattern that this integral fits is the one for the inverse hyperbolic cosine function.

EXAMPLE 9:

SOLUTION:

Here a = 2 and u = x, so the form is a ² - u ², but which formula do we use. Remember that there are two options for this integral. Let us look at the bounds for this integral. When I plug those values in for x, notice that a ² > x ², so you will have to use the inverse hyperbolic tangent function.

EXAMPLE 10:

SOLUTION:

Again a = 2 and u = x, and this integral is of the form a ² - u ², but the bounds of this integral imply that u ² > a ², so I will use the inverse hyperbolic cotangent function.

EXAMPLE 11:

SOLUTION:

Let a = 2 and u = 3x, that means du = 3dx, and this integral is of the form a ² - u ². Therefore, this integral fits the form for the inverse hyperbolic secant function.

EXAMPLE 12:

SOLUTION:

Let a = 5 and u = x, this implies that this integral is of the form a ² + u ². Therefore this integral fits the pattern of the inverse hyperbolic cosecant function.

As I have stated before, there will be a lot of pattern recognition when finding the derivatives of the inverse trig and inverse hyperbolic trig functions. As you work through these examples, please take time to determine the pattern and how it relates to the correct formula. If you have any questions on these examples, please feel free to ask me in my office or via email.

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