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Derivatives of Exponential and Logarithmic FunctionsPowerPoint Presentation

Derivatives of Exponential and Logarithmic Functions

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Derivatives of Exponential and Logarithmic Functions

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Derivatives of Exponential and Logarithmic Functions

Stewart Plus

Find

Find

Because

Use graphing calculator

Therefore: The derivative of f (x) = ex is f ’(x) = ex.

Find f’(x)

- f(x) = 4ex – 8x2 + 7x - 14
f’(x) = 4ex – 16x + 7

- f(x) = x7 – x5 + e3 – x + ex
f’(x) = 7x6 – 5x4 + 0 –1 + ex

= 7x6 – 5x4 –1 + ex

Remember that e is a real number, so the power rule is used to find the derivative of xe.

Also e2 7.389 is a constant, so its derivative is 0.

Find derivatives for

A) f (x) = ex / 2 f ’(x) = ex / 2

B) f (x) = 2ex +x2f ’(x) = 2ex + 2x

C) f (x) = -7xe– 2ex + e2f ’(x) = -7exe-1 – 2ex

is equivalent to

Domain: (0, ∞)

Range: (-∞, ∞)

Range: (0, ∞)

Domain: (-∞, ∞)

* These are inverse function. The graphs are symmetric with respect to the line y=x

* There are many different bases for a logarithmic functions. Two special logarithmic functions are common logarithm (log10x or log x) and natural logarithm (logex = ln x)

1)

2)

3)

4)

5)

Optional slide:

Find

Property 2

Multiply by 1 which is x / x

Set s = h / x

So when h approaches 0, s also approaches o

Property 3

Definition of e

Property 4: ln(e)=1

Therefore: The derivative of f (x) = ln x is f ’(x) =

Find y’ for

A)

B)

The derivative of f(x) = bx

is f’(x) = bx ln b

The derivative of f(x) = logb x

is f’(x) =

Proofs are on page 598

Find g’(x) for

A)

B)

An Internet store sells blankets. If the price-demand

equation is p = 200(0.998)x, find the rate of change of price

with respect to demand when the demand is 400 blankets

and explain the result.

p’ = 200 (.998)x ln(0.998)

p’(400) = 200 (.998)400 ln(0.998) = -0.18.

When the demand is 400 blankets, the price is decreasing about 18 cents per blanket

A model for newspaper circulation is C(t) = 83 – 9 ln t

where C is newspaper circulation (in millions) and t is the number of

years (t=0 corresponds to 1980). Estimate the circulation and find the

rate of change of circulation in 2010 and explain the result.

t = 30 corresponds to 2010

C(30) = 83 – 9 ln30 = 52.4

C(t)’ = C’(30) =

The circulation in 2010 is about 52.4 million and is decreasing at the rate of 0.3 million per year

Y = mx + b

f’(x) = 2ex + 6

m = f’(0) = 2(1) + 6 = 8

y=f (0) = 2(1) + 6(0) = 2

Y = mx + b

2 = 8(0) + b so b = 2

The equation is y = 8x + 2

F(x) = (lnx)2 and g(x) = x

On your calculator, press Y=

Type in the 2 functions above for Y1 and Y2

Press ZOOM, 6:ZStandard

To have a better picture, go back to ZOOM, 2: Zoom In

*Now, to find the point of intersection (there is only 1 in this problem), press 2ND, TRACE then 5: intersect

Play with the left and right arrow to find the linking dot, when you see it, press ENTER, ENTER again, then move it to the intersection, press ENTER. From there, you should see the point of intersection

(.49486641, .49486641)