DEVELOP PROPERTIES OF THE NATURAL EXPONENT FUNCTION

DIFFERENTIATE NATURAL EXPONENTIAL FUNCTIONS

5.4a Exponential Functions: Differentiation

f(x) = ln x is a function that is increasing on its entire domain, and so it has an inverse, f-1(x) = ex

f(x)

f-1(x)

Since they are inverses,ln ex = x and eln x = x

Ex 1 p. 350 Solving Exponential Equations

2Solve 9 xe Take the natural log of both sides (since base e)

2ln 9 ln xe ln 9 2x

2 ln 9 x 0.197 x

Another way to think of this is to put the exponential form into logarithmic form.From the original problem, to free up a variable in the exponential position, rewrite in logarithmic form:

log 9 2e x ln 9 2x 2 ln 9 x 0.197 x

Ex 2 p.351 Solving a logarithmic equationSolve ln(3 5) 8x Exponentiate each side

ln(3 5) 8xe e 83 5x e

83 5x e

995.319x Another way to think of this is to put the log form back into exponential form – the base raised to the exponent(opp. side) is equal to the inside of the log.

83 5x e

8 5

3

ex

And so on!

The coolest thing about ex is that it is its own derivative!

Proof: We know that ln ex =x. So if we take the derivative of both sides, ln xd d

e xdx dx

1

1xx

de

e dx Multiply both sides by ex

x xde e

dx

A geometrical interpretation is that the slope of the graph of f(x) = ex at any point (x, ex) is equal to the y-value of the point.

Ex 3 p. 352 Differentiating Exponential Functions

3 2. xda edx

3 23 xe

2. xdb edx

u = 3x-2

u= -2x-12

2

2 xex

2

2

2 xe

x

Ex 4, p352 Locating Relative Extrema

Find the relative extrema of ( ) 2 xf x xe

Using the product rule, '( ) 2 ( ) (2)x xf x x e e

2 ( 1)xe x

Since is never equal to zero, then x = -1 is the only critical value.xe

In the interval (-∞, -1) the derivative is negative so the function is decreasing. In the interval (-1, ∞), the derivative is positive, so the function is increasing. Since it switches from decreasing to increasing at x = -1, the point (-1, -2/e) is a relative minimum.

Ex 5 p. 353 The Standard Normal Probability Density FunctionShow that the standard normal probability density function

2 21( )

2xf x e

has points of inflection at x = 1

Solution: to locate possible points of inflection, find the x-values for which the second derivative is 0 21

2u x

2 21 2'( )

22xf x e x

2 21

2xx e

2 21''( )

2xd

f x xedx

2 22 21( 1)

2x xx x e e

22 21( 1)( )

2xx e

Setting f “(x) = 0 makes possible points of inflection happen at x = -1, 1Testing intervals using Ch 3 concepts, they are points of inflection when x = 1

5.4a p. 356/ 1-61 every other odd