1. ## Derivatives

How do you do these two derivatives? (Work below.)

1. $\displaystyle y = e^{(1+lnx)}$

2. $\displaystyle y = x^{lnx}$

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1. $\displaystyle y = e^{(1+lnx)}$

$\displaystyle y' = \frac{1}{x}(e^{(1+lnx)})$

Actual answer: $\displaystyle y' = e$

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2. $\displaystyle y = x^{lnx}$

$\displaystyle y' = \frac{1}{x}(lnx)(x^{lnx})$

$\displaystyle y' = \frac{2 (lnx)( x^{lnx})}{x}$

2. Originally Posted by Cursed
How do you do these two derivatives? (Work below.)

1. $\displaystyle y = e^{(1+lnx)}$

2. $\displaystyle y = x^{lnx}$

------------
------------
1. $\displaystyle y = e^{(1+lnx)}$

$\displaystyle y' = \frac{1}{x}(e^{(1+lnx)})$

Actual answer: $\displaystyle y' = e$

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note that this is the answer you have!

recall, $\displaystyle e^{\ln x} = x$. you can use this fact to simplify your answer to get what you have stated.

or you could have done it from the beginning: $\displaystyle e^{1 + \ln x} = e \cdot e^{\ln x} = ex$ ...

2. $\displaystyle y = x^{lnx}$

$\displaystyle y' = \frac{1}{x}(lnx)(x^{lnx})$

$\displaystyle y' = \frac{2 (lnx)( x^{lnx})}{x}$
hint: note that $\displaystyle x^{\ln x} = e^{\ln x^{\ln x}} = e^{\ln x \cdot \ln x} = e^{(\ln x)^2}$

3. Hello, Cursed!

The first one is a trick question . . .

$\displaystyle 1)\;\;y \:=\: e^{1+\ln x}$

We have: .$\displaystyle y \;=\;e^{1+\ln x} \;=\;e^1\cdot e^{\ln x} \;=\;e\cdot x$

Therefore: .$\displaystyle y \:=\:ex \quad\Rightarrow\quad y\:\!' \:=\:e$

$\displaystyle 2)\;\;y \:=\: x^{\ln x}$

This requires Logarithmic Differentiation.

Take logs: .$\displaystyle \ln(y) \;=\;\ln\left(x^{\ln x}\right) \;=\;\ln x\cdot\ln x \quad\Rightarrow\quad \ln(y) \;=\;\left[\ln x\right]^2$

Differentiate implicitly: .$\displaystyle \frac{1}{y}\cdot y\:\!' \;=\;2\cdot\ln x \cdot\frac{1}{x} \quad\Rightarrow\quad \frac{y\:\!'}{y} \;=\;\frac{2\ln x}{x}$

Then: .$\displaystyle y' \;=\;y\cdot\frac{2\ln x}{x} \;=\;\frac{2\,x^{\ln x}\ln x}{x}$