TBIL-Cal Differentiating Inverse Functions (DF8)

whose sine value is given. You can use the unit circle to help you remember which angles yield the given values of sine. The first entry is provided: a sine value of

- 1

corresponds to the angle

- π / 2 .

Table 55.

$y = \sin (x)$	$- 1$	$- \sqrt{3} / 2$	$- 1 / 2$	$0$	$1 / 2$	$\sqrt{3} / 2$	$1$
$x = \sin^{- 1} (y)$	$- π / 2$

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(b)

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From the graph of

y = \sin (x)

and your table above, graph the arcsine function

y = \sin^{- 1} (x)

Figure 56. The graphs of $\sin (x)$ and one point on $\sin^{- 1} (x) .$

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(c)

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Let’s now work with the function arccosine. Again, we need to restrict the domain of cosine to be able to invert the function (Why?). The convention is to restrict cosine to the domain

[0, π]

in order to define arccosine. Given this restriction, what are the domain and range of arccosine? Create a table of values and graph the function arccosine.

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(d)

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Let’s now work with the function arctangent. Again, we need to restrict the domain of tangent to be able to invert the function (Why?). The convention is to restrict tangent to the domain

(- π / 2, π / 2)

in order to define arctangent. Given this restriction, what are the domain and range of arctangent? Create a table of values and graph the function arctangent.

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Activity 2.8.9.

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In this activity you will find a formula for the derivative of arctangent.

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(a)

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Differentiate the implicit equation

\tan (y) = x,

what do you get for

\frac{d y}{d x} ?

$\frac{d y}{d x} = \frac{x}{\tan (y)}$
$\frac{d y}{d x} = \frac{1}{\tan (y)}$
$\frac{d y}{d x} = \frac{x}{\sec^{2} (y)}$
$\frac{d y}{d x} = \frac{1}{\sec^{2} (y)}$

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(b)

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For what function

y = g (x)

have you found the derivative

\frac{d y}{d x} ?

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(c)

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We want to rewrite

\frac{d y}{d x}

only in terms of

x .

Notice that

\tan^{2} (y) = \frac{\sin^{2} (y)}{\cos^{2} (y)} = \frac{1 - \cos^{2} (y)}{\cos^{2} (y)} .

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Multiplying out by the denominator, isolating, and solving for

\cos^{2} (y),

we get that

$\cos^{2} (y) = \frac{\tan^{2} (y)}{\cos^{2} (y)}$
$\cos^{2} (y) = \frac{1}{\tan^{2} (y) + 1}$
$\cos^{2} (y) = \frac{1 - \cos^{2} (y)}{\tan^{2} (y)}$
$\cos^{2} (y) = \frac{1}{\tan^{2} (y) - 1}$

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(d)

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Finally, rewrite

\frac{d y}{d x}

\frac{d y}{d x} = \cos^{2} (y)

and use the fact that

\tan (y) = x

to get a nice formula for the derivative of the arctangent function of

x .

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Remark 2.8.10.

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Consider the functions

y = \tan^{- 1} (x) .

Using your algebra above, you should have found that

\frac{d}{d x} (\tan^{- 1} (x)) = \frac{1}{1 + x^{2}} .

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In a similar fashion, one can find that

\frac{d}{d x} (\sin^{- 1} (x)) = \frac{1}{\sqrt{1 - x^{2}}}, \frac{d}{d x} (\cos^{- 1} (x)) = - \frac{1}{\sqrt{1 - x^{2}}} .

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Activity 2.8.11.

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Demonstrate and explain how to find the derivative of the following functions. Be sure to explicitly denote which derivative rules (product, quotient, sum and difference, etc.) you are using in your work.

$k (t) = \frac{\arctan (- 4 t)}{\ln (- 4 t)}$
$j (u) = - 5 \arcsin (u) \log (u^{6} + 2)$
$n (x) = \ln (- \arcsin (x) + 4 \arctan (x))$

Answer.

$k^{'} (t) = - \frac{\arctan (- 4 t)}{t \log {(- 4 t)}^{2}} - \frac{4}{(16 t^{2} + 1) \log (- 4 t)}$
$j^{'} (u) = - \frac{30 u^{5} \arcsin (u)}{u^{6} + 2} - \frac{5 \log (u^{6} + 2)}{\sqrt{- u^{2} + 1}}$
$n^{'} (x) = \frac{\frac{1}{\sqrt{- x^{2} + 1}} - \frac{4}{x^{2} + 1}}{\arcsin (x) - 4 \arctan (x)}$