Page 88 - Calculus Demystified

CHAPTER 2
SOLUTION Foundations of Calculus 75
Let
2
x
h(x) = ln .
x − 2
Then
h = f ◦ g,
2
where f(s) = ln s and g(x) = x /(x − 2).So (d/ds)f (s) = 1/s and
2
2
2
2
(d/dx)g(x) = (x − 2) · 2x − x · 1/(x − 2) = (x − 4x)/(x − 2) .As a
result,
d d
h(x) = (f ◦ g)(x)
dx dx

df d
= (g(x)) · g(x)
ds dx
2
1 x − 4x
= ·
g(x) (x − 2) 2
2
1 x − 4x
= ·
2
x /(x − 2) (x − 2) 2
x − 4
= .
x(x − 2)
You Try It: Perform the differentiation in the last example by first applying a rule
of logarithms to simplify the function to be differentiated.
x
You Try It: Calculate the derivative of tan(e − x).
EXAMPLE 2.18
Calculate the tangent line to the graph of f(x) = x · e x 2 at the point (1,e).
SOLUTION
The slope of the tangent line will be the derivative. Now

2 2 2
2

f (x) =[x] · e x + x · e x = e x + x · 2x · e x .

In the last derivative we have of course used the Chain Rule. Thus f (1) =
e + 2e = 3e. Therefore the equation of the tangent line is
(y − e) = 3e(x − 1).
You Try It: Calculate the equation of the tangent line to the graph of g(x) =
2
cos((x − 2)/ln x) at the point (2, cos[2/ ln 2]).