But I cant do it :(

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

Printable View

- May 15th 2011, 06:55 AMadam_leedsShould be simple differentation
But I cant do it :(

$\displaystyle F(x) = x{a}^{ 1-x} $ Differentiate with respect to x, so $\displaystyle u = x$ and $\displaystyle v = {a}^{ 1-x}$ But i Cant Differentiate v :( - May 15th 2011, 06:57 AMTheEmptySet
- May 20th 2011, 04:05 AMadam_leeds
- May 20th 2011, 04:57 AMAckbeet
- May 20th 2011, 08:09 AMzebode
- May 20th 2011, 11:18 AMadam_leeds
- May 20th 2011, 11:22 AMAckbeet
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 AMadam_leeds
- May 21st 2011, 06:15 AMadam_leeds
The answer for dv should be $\displaystyle -{a}^{1-x } lna$

They don't look similar (Wondering) - May 21st 2011, 06:33 AMTheEmptySet
$\displaystyle -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) $