Question

(another characterisation of y = e^x -- it is the only exponential function whose gradient at its y-intercept is 1)

(a) prove that y= a^x has derivative at y=a^x loga

(b) prove that y= a^x has gradient 1 at its y intercept if and only if a = e

(c) prove that y=Aa^x has gradient 1 at its y intercept if and only if a=e^(1/a)

Answer 1

f(x) = a^x
f'(x) = [f(x+h) - f(x)]/h
= [a^(x+h) - a^(x)]/h
= [a^(x)a^(h) - a^(x)]/h
= a^(x)[a^(h)-1]/h lim h-> 0
= a^x loga

Answer 2

im alittle confused

Answer 3

why?

Answer 4

is that a general formula for intergrating exponentials?

Answer 5

no...

Answer 6

@and: differentiating.

Answer 7

its the fundamental process for evaluating derivatives..

Answer 8

oh sorry yeah i ment diffrentiating

Answer 9

the part where ive replaced
(a^h - 1 )/ h as loga where h tends to zero is a standard limit definition..

Answer 10

log base e of a right?

Answer 11

yeah

Answer 12

or natural log of a?

Answer 13

cool same page then

Answer 14

ln a

Answer 15

now we want to show the slope is 1 at the y-intercept( or when x=0) only if a=1.
f'(0)=a^0 *lna=lna
lna=1 only when a=e.

Answer 16

only if a=e

Answer 17

*

Answer 18

is there n easier proof because i havent learnt the general formula you used

Answer 19

it depends what you can use to prove
can use that the derivative of lny =y'/y

Answer 20

yeah ive seen that before

Answer 21

y=a^x
lny=lna^x
lny=xlna
y'/y=lna
y'=ylna
y'=a^x*lna

Answer 22

oh i see. i understand now thanks

Answer 23

but what him mentioned above his the fundamental definition of derivative and you have not seen it?
it just means as the slopes of secants gets closer to the point of tangency we can find the slope of the tangent line

Answer 24

but you haven't seen that form there is another form
f'(x0)=limx->x0[(f(x)-f(x0))/(x-x0)]

Answer 25

no i didnt understand the last line of his one

Answer 26

the third question is that a=e^(1/A)

Answer 27

do you differentiate it then sub in a=e^(1/A) to see if it is true?

Answer 28

so we have
y=Aa^x
we already found that if y=a^x then y'=a^x*lna
so if y=Aa^x then y'=Aa^xlna
yes just put where there is an a, e^(1/A)
also y-intercept means x=0
so we have
y'=Aa^0*lna=Alna=Aln(e^[1/A])=A*1/A*lne=1*1=1

Answer 29

ok thanks heaps :)