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Proof of the differential of a^x = a^xlna Watch

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    Need a clear explanation please so that I can figure out the differential of x = 2^t and y = x^x.
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    Well, say you have y=a^x,
    Take logs of both sides,
    lny=ln(a^x)=xlna
    then y=e^(xlna)=a^x
    Then differentiate that normally, i.e. f'(x)e^[f(x)]
    then y'=lna.e^(xlna)=lna.a^x
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    (Original post by Toneh)
    Well, say you have y=a^x,
    Take logs of both sides,
    lny=ln(a^x)=xlna
    then y=e^(xlna)=a^x
    Then differentiate that normally, i.e. f'(x)e^[f(x)]
    then y'=lna.e^(xlna)=lna.a^x
    So you've said that the differential of xlna is lna (which is what you multiply e^xlna by), whereas product rule says that the differential of xlna is x/a + lna? Not sure I've followed you. How does xlna differentiate to lna?
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    (Original post by A level Az)
    So you've said that the differential of xlna is lna (which is what you multiply e^xlna by), whereas product rule says that the differential of xlna is x/a + lna? Not sure I've followed you. How does xlna differentiate to lna?
    a is a constant, therefore so is lna, so product rule would yield lna.dx/dx + x.d(lna)/dx = lna + 0
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    (Original post by Toneh)
    a is a constant, therefore so is lna, so produce rule would yield lna.dx/dx + x.d(lna)/dx = lna + 0
    Ah okay, so if I just think of it as ln(a).x instead of xlna then I should remember that lna is a constant Thanks.
 
 
 
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