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

1. Need a clear explanation please so that I can figure out the differential of x = 2^t and y = x^x.
2. 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
3. (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?
4. (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
5. (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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