Should be simple differentation

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May 15th 2011, 06:55 AM
adam_leeds

Should be simple differentation

But I cant do it :(

$F(x) = x{a}^{ 1-x}$ Differentiate with respect to x, so $u = x$ and $v = {a}^{ 1-x}$ But i Cant Differentiate v :(
May 15th 2011, 06:57 AM
TheEmptySet

Quote:

Originally Posted by adam_leeds

But I cant do it :(

$F(x) = x{a}^{ 1-x}$ Differentiate with respect to x, so $u = x$ and $v = {a}^{ 1-x}$ But i Cant Differentiate v :(

use algebra

$v=a\cdot a^{-x}=ae^{\ln(x^{-1})}= ae^{-\ln(x)}$

Can you finish from here?
May 20th 2011, 04:05 AM
adam_leeds

Quote:

Originally Posted by TheEmptySet

use algebra

$v=a\cdot a^{-x}=ae^{\ln(x^{-1})}= ae^{-\ln(x)}$

Can you finish from here?

$\frac{-a}{ x} {e}^{ -ln(x)}$

?
May 20th 2011, 04:57 AM
Ackbeet

Quote:

Originally Posted by TheEmptySet

use algebra

$v=a\cdot a^{-x}=ae^{\ln(x^{-1})}= ae^{-\ln(x)}$

Can you finish from here?

I think there is an algebra error here. I get

$v=a\cdot a^{-x}=a\,e^{\ln(a^{-x})}=a\,e^{-x\ln(a)}}.$
May 20th 2011, 08:09 AM
zebode

Quote:

Originally Posted by Ackbeet

I think there is an algebra error here. I get

$v=a\cdot a^{-x}=a\,e^{\ln(a^{-x})}=a\,e^{-x\ln(a)}}.$

I think too.. I get this same thing...
May 20th 2011, 11:18 AM
adam_leeds

Quote:

Originally Posted by Ackbeet

I think there is an algebra error here. I get

$v=a\cdot a^{-x}=a\,e^{\ln(a^{-x})}=a\,e^{-x\ln(a)}}.$

How do you differentiate that wrt to x?
May 20th 2011, 11:22 AM
Ackbeet

Well, the same way you'd differentiate exp(ax): by using the chain rule. The expression -ln(a) is just a constant, right?
May 21st 2011, 06:11 AM
adam_leeds

Quote:

Originally Posted by Ackbeet

Well, the same way you'd differentiate exp(ax): by using the chain rule. The expression -ln(a) is just a constant, right?

Yeah

so it is $-aln(a){e}^{-xln(a) }$
May 21st 2011, 06:15 AM
adam_leeds

The answer for dv should be $-{a}^{1-x } lna$

They don't look similar (Wondering)
May 21st 2011, 06:33 AM
TheEmptySet

$-a \ln(a)e^{-x\ln(a)}=-a \ln(a)e^{\ln(a^{-x})}= -a\ln(a) \cdot a^{-x}=-a^{1-x}\ln(a)$

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