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anonymous

  • 5 years ago

(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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  1. anonymous
    • 5 years ago
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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

  2. anonymous
    • 5 years ago
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    im alittle confused

  3. anonymous
    • 5 years ago
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    why?

  4. anonymous
    • 5 years ago
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    is that a general formula for intergrating exponentials?

  5. anonymous
    • 5 years ago
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    no...

  6. anonymous
    • 5 years ago
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    @and: differentiating.

  7. anonymous
    • 5 years ago
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    its the fundamental process for evaluating derivatives..

  8. anonymous
    • 5 years ago
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    oh sorry yeah i ment diffrentiating

  9. anonymous
    • 5 years ago
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    the part where ive replaced (a^h - 1 )/ h as loga where h tends to zero is a standard limit definition..

  10. myininaya
    • 5 years ago
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    log base e of a right?

  11. anonymous
    • 5 years ago
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    yeah

  12. myininaya
    • 5 years ago
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    or natural log of a?

  13. myininaya
    • 5 years ago
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    cool same page then

  14. anonymous
    • 5 years ago
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    ln a

  15. myininaya
    • 5 years ago
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    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.

  16. myininaya
    • 5 years ago
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    only if a=e

  17. myininaya
    • 5 years ago
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    *

  18. anonymous
    • 5 years ago
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    is there n easier proof because i havent learnt the general formula you used

  19. myininaya
    • 5 years ago
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    it depends what you can use to prove can use that the derivative of lny =y'/y

  20. anonymous
    • 5 years ago
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    yeah ive seen that before

  21. myininaya
    • 5 years ago
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    y=a^x lny=lna^x lny=xlna y'/y=lna y'=ylna y'=a^x*lna

  22. anonymous
    • 5 years ago
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    oh i see. i understand now thanks

  23. myininaya
    • 5 years ago
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    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

  24. myininaya
    • 5 years ago
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    but you haven't seen that form there is another form f'(x0)=limx->x0[(f(x)-f(x0))/(x-x0)]

  25. anonymous
    • 5 years ago
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    no i didnt understand the last line of his one

  26. myininaya
    • 5 years ago
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    the third question is that a=e^(1/A)

  27. anonymous
    • 5 years ago
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    do you differentiate it then sub in a=e^(1/A) to see if it is true?

  28. myininaya
    • 5 years ago
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    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

  29. anonymous
    • 5 years ago
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    ok thanks heaps :)

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