(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)

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(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)

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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
im alittle confused
why?

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is that a general formula for intergrating exponentials?
no...
@and: differentiating.
its the fundamental process for evaluating derivatives..
oh sorry yeah i ment diffrentiating
the part where ive replaced (a^h - 1 )/ h as loga where h tends to zero is a standard limit definition..
log base e of a right?
yeah
or natural log of a?
cool same page then
ln a
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.
only if a=e
*
is there n easier proof because i havent learnt the general formula you used
it depends what you can use to prove can use that the derivative of lny =y'/y
yeah ive seen that before
y=a^x lny=lna^x lny=xlna y'/y=lna y'=ylna y'=a^x*lna
oh i see. i understand now thanks
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
but you haven't seen that form there is another form f'(x0)=limx->x0[(f(x)-f(x0))/(x-x0)]
no i didnt understand the last line of his one
the third question is that a=e^(1/A)
do you differentiate it then sub in a=e^(1/A) to see if it is true?
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
ok thanks heaps :)

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