Page 87 - Calculus Demystified

CHAPTER 2
74
Therefore, by the quotient rule, Foundations of Calculus
x x x
d e + x sin x tan x · (d/dx)(e + x sin x) − (e + x sin x)(d/dx) tan x
=
dx tan x (tan x) 2
x
x
tan x · (e + sin x + x cos x) − (e + x sin x) · (sec x) 2
=
(tan x) 2
x
x
2
2
e tan x + tan x sin x + x sin x − e sec x − x sin x sec x
= .
2
tan x

d x
You Try It: Calculate the derivative sin x · cos x − .
x
dx e + ln x
EXAMPLE 2.16
Calculate the derivative
d 3 2
(sin(x − x )).
dx
SOLUTION
This is the composition of functions, so we must apply the Chain Rule. It is
essential to recognize what function will play the role of f and what function
will play the role of g.
3
2
Notice that, if x is the variable, then x − x is applied first and sin
3
2
applied next. So it must be that g(x) = x − x and f(s) = sin s. Notice that
2
(d/ds)f (s) = cos s and (d/dx)g(x) = 3x − 2x. Then
2
3
sin(x − x ) = f ◦ g(x)
and
d 3 2 d
(sin(x − x )) = (f ◦ g(x))
dx dx

df d
= (g(x)) · g(x)
ds dx
2
= cos(g(x)) · (3x − 2x)
2
3
2
=[cos(x − x )]· (3x − 2x).
EXAMPLE 2.17
Calculate the derivative
2
d x
ln .
dx x − 2