MATH241 Lecture Notes - Lecture 19: Relative Growth Rate
9 May 2018
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MATH241 - Lecture 19 - Derivatives and Exponential Growth and Decay
Derivatives involving Functions with Variable Exponents
Case I: Constant base with a variable exponent
Let y=bx
y=eln bx
y=ex ln b
n b ey′=lx ln b
= (ln b)bx
ln b=bx
Example
:
Differentiate
A) y= 2x
B) y= 2x2
Solution:
A) ln 2y′= 2x
B) y′= 2 ln 2
(x3)3x
(2)
2ln 2y′= 3x2x3
Case II: Variable base, variable exponent
y=xcos x
n y n xl =lcos x
os x ln x=cd
dx =x
1
in x ln x
y
y′= − s+x
cos x
y′=y−in x ln x
(s+x
cos x)
y′=xcos x−in x ln x
(s+x
cos x)
Document Summary
Math241 - lecture 19 - derivatives and exponential growth and decay. Case i: constant base with a variable exponent. Let y = bx y = eln bx y = ex ln b n b e y = l x ln b. Differentiate: y = 2x, y = 2x2. B) y = 2 ln 2 y = 2x ln 2 ( x3 y = 3x2 x3. Case ii: variable base, variable exponent y = xcos x n x l n y cos x os x ln x. = c y = s cos x. + x y ( s cos x) y = y in x ln x. + x cos x) y = xcos x in x ln x. + x in x ln x ( s d. If the rate of change of y in respect to t is proportional to its size, then dy = l dt n y. Recall that if y = c kt e.
