MA 242 Review Exponential and Log Functions Notes for today s class can be found at www.xecu.net/jacobs/index242.htm
Example: If y = x n If y = x 2 then then dy dx = nxn 1 dy dx = 2x1 = 2x
Power Function
Exponential Function
Chromium-254 is a radioactive isotope with a half-life of approximately 1 month.
Chromium-254 is a radioactive isotope with a half-life of approximately 1 month. Let s suppose we begin with an 8 gram sample of chromium. After 1 month, we will be left with a 4 gram sample
Number of months Mass of chromium 0 8 grams 1 4 grams 2 2 grams 3 1 gram 4 1 2 gram
Number of months Mass of chromium 0 8 grams 1 8 1 2 = 4 grams 2 8 1 2 1 2 = 2 grams 3 8 1 2 1 2 1 2 = 1 gram 4 8 1 2 1 2 1 2 1 2 = 1 2 gram
Number of months Mass of chromium 0 8 grams 1 8 1 2 = 4 grams 2 8 ( 1 2) 2 = 2 grams 3 8 ( 1 3 2) = 1 gram 4 8 ( ) 1 4 2 = 1 2 gram ( ) t 1 f(t) = 8 2
f(t) = 8 ( ) t 1 2
Giardia
The number of giardia organisms will quadruple every day
Number of days 0 1 1 4 2 4 2 = 16 3 4 3 = 64 4 4 4 = 256 f(t) = 4 t Number of organisms
f (t) = 4t
Derivative of a Power Function: d dx (xn ) = nx n 1 Derivative of an Exponential Function: d dx (ax ) =????
f(x) = a x f f(x + h) f(x) (x) = lim h 0 h a x+h a x = lim h 0 h
f(x) = a x f f(x + h) f(x) (x) = lim h 0 h a x+h a x = lim h 0 h a x a h a x = lim h 0 h = a x a h 1 lim = a x k h 0 h
Derivative of a Power Function: d dx (xn ) = nx n 1 Derivative of an Exponential Function: d dx (ax ) = k a x
d dx (ax ) = k a x d dx (2x ) = (0.693)2 x d dx (3x ) = (1.099)3 x
d dx (2x ) = (0.693)2 x d dx (3x ) = (1.099)3 x d dx (ex ) = e x e = 2.71828...
y = e x dy dx = ex
Problem: If y = e sin x, find dy dx
Problem: If y = e sin x, find dy dx Let u = sin x so y = e u. According to the Chain Rule, dy dx = dy du du dx = eu du dx = esin x cos x
Generalization: d dx ( e f(x)) = e f(x) f (x)
Properties of exponential 1. e 0 = 1 2. e 1 = e 3. e u+v = e u e v 4. e u v = eu e v 5. (e u ) n = e nu d 6. dx (ex ) = e x ( ) e f(x) = e f(x) f (x) 7. d dx
Let x > 0 y = log b x means b y = x
ln x means log e x so y = ln x implies e y = x
Properties of exponential and logarithms 1. e 0 = 1 ln 1 = 0 2. e 1 = e ln e = 1 3. e u+v = e u e v ln(xy) = ln x + ln y 4. e u v = eu e ln x v y = ln x ln y 5. (e u ) n = e nu ln (x n ) = n ln x ( d 6. ) dx e f(x) = e f(x) f d (x) dx (ln x) = 1 x
If x > 0 then: d dx (ln x) = 1 x 1 dx = ln x + C x
Derivative Example: Let y = sin(ln x) Find dy dx
Let u = ln x so y = sin u Let y = sin(ln x) Find dy dx dy dx = dy du du dx = cos u 1 x = 1 x cos(ln x)
Similar Derivative Example: Let y = ln(sin x) Find dy dx
Let u = sin x so y = ln u Let y = ln(sin x) Find dy dx dy dx = dy du du dx = 1 u du dx = 1 sin x cos x = cot x
More generally, suppose: y = ln f(x) Find the derivative
y = ln f(x) Let u = f(x) so y = ln u dy dx = dy du du dx = 1 u du dx = 1 f(x) f (x)
Example: Find the derivative d 1 (ln f(x)) = dx f(x) f (x) y = ln ( x ) dy dx = 1 x d dx ( x)
Find the derivative y = ln ( x ) dy dx = 1 x 1/2 d dx (x 1/2) = x 1/2 1 2 x 1/2 = 1 2 x 1 = 1 2x
Example: Find the derivative y = ln ( x ) Alternate solution: dy dx = d dx y = ln ( x ) = ln ( ) 1 2 ln x = 1 2 ( x 1/2) = 1 2 ln x d dx (ln x) = 1 2 1 x = 1 2x
Find the derivative of: y = ln ( x 2 cos x )
y = ln ( x 2 cos x ) = ln ( x 2) + ln cos x = 2 ln x + ln cos x dy dx = 2 1 x + 1 cos x ( sin x) = 2 x tan x
Find the derivative of: y = ln(sec x + tan x)
Find the derivative of: y = ln(sec x + tan x) dy dx = 1 sec x + tan x d (sec x + tan x) dx
Find the derivative of: dy dx = 1 sec x + tan x y = ln(sec x + tan x) d (sec x + tan x) dx 1 = sec x + tan x (sec x tan x + sec2 x) 1 = (tan x + sec x) sec x sec x + tan x = sec x
Similarly d (ln(sec x + tan x)) = sec x dx sec x dx = ln(sec x + tan x) + C d (ln(csc x + cot x)) = csc x dx csc x dx = ln(csc x + cot x) + C
More generally, d dx (ln x) = 1 x d dx (ln x ) = 1 x d dx (ln f(x)) = f (x) f(x) d dx (ln f(x) ) = f (x) f(x)
d dx (ln x ) = 1 x d dx (ln f(x) ) = f (x) f(x) 1 dx = ln x + C x f (x) dx = ln f(x) + C f(x)
1 2x + 1 dx
Let u = 2x + 1 du dx = 2 1 2du = dx 1 1 2x + 1 dx = u 1 2 du = 1 1 2 u du = 1 2 ln u + C = 1 ln 2x + 1 + C 2
Alternate approach: Make use of the formula: f (x) f(x) dx = ln f(x) + C 1 2x + 1 dx = 1 2 2 2x + 1 dx = 1 2 ln 2x + 1 + C
1/2 0 x 2x + 1 dx
1/2 0 x 2x + 1 dx Let u = 2x + 1 x = 1 2 (u 1) dx = 1 2 du 1/2 0 x 2x + 1 dx = = 1 4 2 1 2 1 1 2 (u 1) 2 du u 1 = 1 ln 2 4 ( 1 1 u ) du
tan x dx
tan x dx = sin x cos x dx
Let t = cos x dt = sin x dx dt = sin x dx sin x tan x dx = cos x dx 1 = t dt = ln t + C = ln cos x + C
e x e x e x dx + e x
Let t = e x + e x dt = (e x e x ) dx e x e x e x dx = + e x 1 t dt = ln t + C = ln ( e x + e x) + C
Note: 1 2 (ex e x ) is called a hyperbolic sine and 1 2 (ex + e x ) is called a hyperbolic cosine sinh x = 1 2 ( e x e x) cosh x = 1 2 ( e x + e x)
sinh x = 1 2 ( e x e x) cosh x = 1 2 ( e x + e x) tanh x = sinh x 1 cosh x = 2 (ex e x ) 1 2 (ex + e x ) = ex e x e x + e x e x e x e x dx = tanh x dx + e x