For the question e^(xlnx), the derivative is given as x^x(1+lnx). I understand where the (1+lnx) comes from, that is the derivative of xlnx, but I'm not sure why it's being multiplied by x^x. I thought that d/dx(e^u) = u' e^u

Second, for the question y=lnx at the point where x=e/2, find the equation of the tangent at that point. Isn't the derivative and slope here 1/2e? The equation given as the answer is y = (2/e)x

Originally Posted by theowne

For the question e^(xlnx), the derivative is given as x^x(1+lnx). I understand where the (1+lnx) comes from, that is the derivative of xlnx, but I'm not sure why it's being multiplied by x^x. I thought that d/dx(e^u) = u' e^u

you are correct in your last formula.

so doing it directly, you would obtain, $\displaystyle (1 + \ln x)e^{x \ln x}$

but you could also realize that $\displaystyle e^{x \ln x} = e^x e^{\ln x} = xe^x$

and find the derivative of that by the product rule

Second, for the question y=lnx at the point where x=e/2, find the equation of the tangent at that point. Isn't the derivative and slope here 1/2e? The equation given as the answer is y = (2/e)x

the slope is not what you say

the derivative is 1/x, if you plug in x = e/2, you get the slope being 2/e

For the first one, note that $\displaystyle e^{xlnx} = e^{{lnx}^x}$.

Recall that $\displaystyle e^{lnx} = x$ and you get $\displaystyle x^x$.

Originally Posted by Jhevon

$\displaystyle (1 + \ln x)e^{x \ln x}$

but you could also realize that $\displaystyle e^{x \ln x} = e^x e^{\ln x} = xe^x$

No: $\displaystyle e^x e^{\ln x} = e^{x + lnx}$, not $\displaystyle x^{xlnx}$

Originally Posted by Mathnasium

No: $\displaystyle e^x e^{\ln x} = e^{x + lnx}$, not $\displaystyle x^{xlnx}$

correct. sorry about that. was a little distracted.

i still think the formula the asker posted first is the easier way though

$\displaystyle e^{x\ln (x)} = e^{\ln \left( {x^x } \right)} = x^x \,\& \,y = x^x \Rightarrow \quad y' = x^x \left( {1 + \ln (x)} \right)
$

Originally Posted by Plato

$\displaystyle e^{x\ln (x)} = e^{\ln \left( {x^x } \right)} = x^x \,\& \,y = x^x \Rightarrow \quad y' = x^x \left( {1 + \ln (x)} \right)
$

just to add. Plato used logarithmic differentiation here