Question

I need help with my Application of Differentiation homework, Anyone want to help me? Please do!! :(

Answer 1

xlnx-2x x>0

Answer 2

whats the question?

Answer 3

xlnx-2x x>0

Answer 4

oh you have to find the coordinates of the min and max points

Answer 5

is it xlnx -2x ; x>0

Answer 6

then differentiate the given expression. At maxima and minima the df/dx =0

Answer 7

yeah but how do I differentiate xlnx

Answer 8

oh don't you know how to differentiate product functions:
d(f(x)g(x))/dx = f(x)g'(x) + g(x)f'(x)

Answer 9

x d(lnx)/dx + lnx dx/dx

Answer 10

To find dlnx/dx;
Assume lnx =y ; So dy/dx is what u want.......right?
or i can write x = e^y
or dx/dx = d(e^y)/dx
or 1 = {d(e^y)/dy}*dy/dx

Answer 11

I guess u can take the work up frm here :P

Answer 12

so.. do I differentiate xlnx and -2x seperately and add them or multiply them.. ? :

Answer 13

.differentiation confuses me and so that makes my homework tonight more difficult, but I don't have time to go back over differentiation cause I find it hard and I've a load to do.

Answer 14

when differentiation acts on sum of functions it is quite simple:
d(f(x)+g(x))/dx =f'(x) + g'(x)

Answer 15

d(xlnx -2x)/dx = d(xlnx)/dx -d(2x)/dx

Answer 16

@yaysocks : don't worry you will get used to it.......everyone becomes comfortable after a point

Answer 17

I'm sorry, I know I'm completely clueless, but now I have lnx=1, how do I get the turning point(s)?

Answer 18

Thank you for your patience!

Answer 19

and help of course!