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    Does e^x tend toward infinity or zero?
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    (Original post by geeeek)
    Does e^x tend toward infinity or zero?
    Depends... what does x tend to?
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    (Original post by RDKGames)
    Depends... what does x tend to?

    Does it tend towards infinity since the curve goes upwards
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    (Original post by geeeek)
    Does it tend towards infinity since the curve goes upwards
    Again... depends what your x tends to...

    BUT I'm gonna shoot in the dark here and say you're wandering what e^x tends to as x \rightarrow \infty

    In which case, yes it does tend to infinity, but not for that reason explicitly.
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    (Original post by RDKGames)
    Again... depends what your x tends to...

    BUT I'm gonna shoot in the dark here and say you're wandering what e^x tends to as x \rightarrow \infty

    In which case, yes it does tend to infinity, but not for that reason explicitly.

    Yes it was regarding e^x in general. Thank you!

    Also, is it possible to differentiate 3^x ?
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    (Original post by geeeek)
    Yes it was regarding e^x in general. Thank you!

    Also, is it possible to differentiate 3^x ?
    Yes it is.
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    (Original post by geeeek)
    Yes it was regarding e^x in general. Thank you!

    Also, is it possible to differentiate 3^x ?
    It is possible, yes. But it is not as straight forward.

    One possible method is to rewrite 3^x as (e^ln3)^x

    Hence 3^x = e^(xln3)

    We are familiar with differentiating e^x and lnx with respect to x so we can easily differentiate the above expression.

    d/dx [e^(xln3)] = ln3 * e^(xln3) = ln(3) * 3^x

    Hence d/dx 3^x = ln(3) * 3^x
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    (Original post by RDKGames)
    Yes it is.
    how would u go about differentiating it?

    will it become x3^x-1 ???
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    (Original post by geeeek)
    how would u go about differentiating it?

    will it become x3^x-1 ???
    Say 3^x=e^{\ln(3^x)}=e^{x\ln(3)} then you have something of the form e^{ax} to differentiate.
 
 
 
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