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)

Results 1 to 6 of 6

- Nov 11th 2007, 02:24 PM #1

- Joined
- Nov 2007
- Posts
- 7

- Nov 11th 2007, 02:34 PM #2
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)

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)

- Nov 11th 2007, 02:36 PM #3

- Joined
- Nov 2007
- Posts
- 7

- Nov 11th 2007, 02:40 PM #4

- Joined
- Nov 2007
- Posts
- 7

- Nov 11th 2007, 02:44 PM #5
you can simplify that so much more

EDIT:

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

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

and remember, ln 10 is a constant

- Nov 11th 2007, 02:45 PM #6