# Math Help - Should be simple differentation

1. ## 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

2. 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?

3. Originally Posted by TheEmptySet
use algebra

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

Can you finish from here?
$\frac{-a}{ x} {e}^{ -ln(x)}$

?

4. 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

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

5. 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...

6. 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?

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

8. 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 $-aln(a){e}^{-xln(a) }$

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

They don't look similar

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