# Math Help - Derivative of e^(xlnx)

1. ## Derivative of e^(xlnx)

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

2. 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, $(1 + \ln x)e^{x \ln x}$

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

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

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

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

4. Originally Posted by Jhevon
$(1 + \ln x)e^{x \ln x}$

but you could also realize that $e^{x \ln x} = e^x e^{\ln x} = xe^x$
No: $e^x e^{\ln x} = e^{x + lnx}$, not $x^{xlnx}$

5. Originally Posted by Mathnasium
No: $e^x e^{\ln x} = e^{x + lnx}$, not $x^{xlnx}$
correct. sorry about that. was a little distracted.

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

6. $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)
$

7. Originally Posted by Plato
$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