Derivative of e^(xlnx)

**theowne** · Mar 30th 2008, 01:01 PM

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

**Jhevon** · Mar 30th 2008, 01:09 PM

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

**Mathnasium** · Mar 30th 2008, 01:11 PM

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

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

**Mathnasium** · Mar 30th 2008, 01:13 PM

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 $x^{xlnx}$

**Jhevon** · Mar 30th 2008, 01:14 PM

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

**Plato** · Mar 30th 2008, 01:55 PM

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

**Jhevon** · Mar 30th 2008, 09:38 PM

Originally Posted by Plato

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

just to add. Plato used logarithmic differentiation here

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