Calculus: Analytic Geometry and Calculus, with Vectors

9.2 Derivatives and integrals of exponentials and logarithms 501

tive, we can begin by being irked by the fact that the base is not e. There are

two superficially different ways to put the given equation in a form to which

basic rules are applicable. Taking logarithms (with base e) gives

(1) log Y = g(x) log f(x)

and, with or without using (1), we can put the equation in the form

(2) y = ea(x> log 1(z).

When (1) is used, we differentiate to obtain

(3)

1

ydx = g(x) fix)f'(x) + g,(X) log f(x)

and, with the aid of the fact that y = f(x)o(x>,

(4) dx

f(x) [

f(x) + g ,(X)logf(x)].

When (2) is used, we differentiate to obtain

(5) dx=ef(x)log 0(x)[f(x)

g 1 g'(x) + f'(x)logg(x) ]

and then use (2) to obtain the result (4). With the above formulas out of sight,

use one of the above ideas to obtain a formula for dy/dx when

Y=xx

(b) y = (1 + x)'Ix

(c) y = (sin x)°ln x

(d) y = (sin x)tsn x

(e) y = xlog x

(f)Y=ab=

(g) Y = x",

flns.: dz = xx(1 + log x)

dy x - (1 + x) log (I + x)

lIns.:dx= (1 + x)llx

x-(1 + x)

flns.: dx = cos x(1 + log sin x)(sin x)-in

Ins.: dx = (1 + sec2 x log sin x) (sin

Ans.: dx = 2x1og z-1 log x

As.:

dy= abobz log b log a

1Qns.:dx=xbbxlog x[ logb+xl 1 -

16 One who is interested in solving a puzzle with an interesting answer may
undertake to determine the nature of the graph of the equation y = xz. One
who has never thought about the orders of magnitudes of the numbers (0.01)0 01
and (j)3h may even be surprised by the results.
17 If the result has not already been obtained, let y = xx and show that

y"(x) = xx(1 + log x)2 + xx-> > 0.

Tell what this says about the graph of y = xx.