If we differentiate both sides of this equation, we get

i.e.

so by the chain rule

(Here I have used the result of exercise 14.37.) Thus the formula

which we have known for quite a while for rational exponents, is actually valid for all real exponents.

Hence

so has an inverse function which is denoted by . By the inverse function theorem is differentiable on . and we have

By the chain rule

Now since the sine function is positive on we get

for all , so

Thus

Show that has an inverse function that is differentiable on the interior of its domain. This inverse functions is called . Describe the domain of , sketch the graphs of and of , and show that

Then T is continuous, and the image of is unbounded both above and below, so image() =

so has an inverse function, which we denote by .

so by the chain rule

Now

so

Thus

Show that has an inverse function arccot, and that

What is ? Sketch the graphs of and of arccot.

**Remark** The first person to give a name to the
inverse trigonometric functions was Daniel Bernoulli (1700-1792)
who used for in 1729. Other early notations included
arc(cos. = ) and ang(cos. = )[15, page 175].
Many calculators and some calculus books use to denote
arccos. (If you use your calculator to find inverse trigonometric
functions, make sure that you set the degree-radian-grad mode to radians.)

- a)
- .
- b)
- .
- c)
- .
- d)
- .
- e)
- .
- f)
- .
- g)
- .

Calculate the derivative of . What is the domain of this function? Sketch the graph of .

These functions are called the

and

Calculate

and simplify your answer as much as you can. What conclusion can you draw from your answer? Sketch the graphs of and on one set of coordinate axes.