The functions and are both defined on the interval

MATHEMATICS · Multi-part · AP · Calculus AB · Differentiation: Composite, Implicit, and Inverse Functions · Differentiation: Composite, Implicit, and Inverse Functions

Problem

The functions

y=\arcsin x

and

y=\arccos x

are both defined on the interval

(-1,1)

, but one is increasing while the other is decreasing. (a) Without using explicit differentiation, describe geometrically why the slope of

y=\arcsin x

is positive and the slope of

y=\arccos x

is negative for every

x

(-1,1)

. Base your explanation on the shapes and orientations of their graphs. (b) From your reasoning in part (a), write an inequality comparing

\dfrac{d}{dx}\arcsin x

and

\dfrac{d}{dx}\arccos x

for

x\in(-1,1)

. (c) Let

u(x)

be a differentiable function with

|u(x)|<1

for all

x

in an interval

I

. Explain why

\dfrac{d}{dx}\arccos\bigl(u(x)\bigr)

differs from

\dfrac{d}{dx}\arcsin\bigl(u(x)\bigr)

by a negative sign, and discuss the role of the Chain Rule in this conclusion.

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