1. ## y'

am i doing this right, and how to make the answer more compact:
1: y=x-x*ln(x)
y'= -x*(1/x) + -ln(x) = -x/x + -ln(x)

2: y=xe^(2x)
y'=x*2e^(2x)+e^2(x)

3: y=3x*log(x)
y'= 3x*1/(x*ln(10))+3*log(x) = 3x/(x*ln(10))+3*log(x) = 3/ln(10)+3*log(x)

2. Originally Posted by stekemrt
am i doing this right, and how to make the answer more compact:
1: y=x-x*ln(x)
y'= -x*(1/x) + -ln(x) = -x/x + -ln(x)
this is wrong

use the product rule and remember you have an x in front where the derivative is 1 (and also, -x/x = -1)

2: y=xe^(2x)
y'=x*2e^(2x)+e^2(x)
this is fine. to get a more compact answer, factor out the e^(2x)

3: y=3x*log(x)
y'= 3x*1/(x*ln(10))+3*log(x) = 3x/(x*ln(10))+3*log(x) = 3/ln(10)+3*log(x)
this is also wrong. remember that $\displaystyle \log_{10}x = \frac {\ln x}{\ln 10}$ and use the product rule

3. ok,
1: y'=1-1*ln(x)+-x*(1/x) = -ln(x)

right?

EDIT:
3: y'=3*log(x)+3x*1/(x*ln(10)) = 3*log(x)+3/(ln(10))

4. Originally Posted by Jhevon
this is fine. to get a more compact answer, factor out the e^(2x)
i dont know how to do that

5. Originally Posted by stekemrt
ok,
1: y'=1-1*ln(x)+-x*(1/x)

right?
you can simplify that so much more

EDIT:
3: y'=3*log(x)+3x*1/(x*ln(10)) = 3*log(x)+3/(ln(10))
i have no idea what you did. let me state it explicitly, what you want is to find the derivative of:

$\displaystyle y = 3x \cdot \frac {\ln x}{\ln 10} = \frac 3{\ln 10}x \ln x$

and remember, ln 10 is a constant

6. Originally Posted by stekemrt
i dont know how to do that
$\displaystyle e^{2x}(2x + 1)$