Question

whats the derivative 2y(ln(x)) and the derivative of 3x(ln(y))

Answer 1

implicits you mean?

Answer 2

yesss

Answer 3

[ 2y(ln(x))]' = 2y'(ln(x))+2y(ln(x))'
= 2 y' (ln(x))+ 2y/x
-2y/x = 2 y' (ln(x))
-2y/2xln(x) = y'

-y/xln(x) = y'

Answer 4

id say its the same wrap up for the other one as well

Answer 5

but i read it wrong lol

Answer 6

wait i got confused..

Answer 7

[3x(ln(y))]' = 3x(ln(y)) + 3x(ln(y))
= 3x'(ln(y)) + 3x(ln(y))'
= 3(ln(y)) + 3x/y y'
-3(ln(y)) = 3x/y y'
-3(ln(y))/3x = 1/y y'
(ln(y))/x = 1/y y'

y(ln(y))/x = y'

Answer 8

which part?

Answer 9

ok o those two sets are suppose to equal each other and im suppose to find y'

but on the first one where you put +2y/x -2y/x...whered you get te -2y/x

Answer 10

first set:

[ 2y(ln(x))]' = 2y'(ln(x))+2y(ln(x))' ; this is just the product rule

2y derives to 2 y'
ln(x) derives to 1/x x', but x'=1 here (dx/dx)

= 2 y' (ln(x)) + 2y/x
right?

Answer 11

yeaa. i got 2y'lnx+2y/x

Answer 12

ok, now solve for y'

Answer 13

-2y/x and divide of that 2lnx

Answer 14

wait noo. 2ylnx=3xlny lol so i look for y' out of that

Answer 15

ok so for the right side you got what? i got 3lny+3xy'/y

Answer 16

is this what your trying to derive?
\[2y\ ln(x) = 3x\ ln(y)\]

Answer 17

yessir

Answer 18

lol ... youre hiding things form me :)

\[\frac{d}{dx}(2y\ ln(x) = 3x\ ln(y))\]
\[2y'\ ln(x)+2y\ ln(x)' = 3x'\ ln(y)+3x\ ln(y)'\]
\[y'\ 2ln(x)+\frac{2y}{x} = 3ln(y)+\frac{3x}{y}\ y'\]
\[y'\ 2ln(x) -\frac{3x}{y}\ y'= 3ln(y)-\frac{2y}{x}\]
\[y'\ (2ln(x) -\frac{3x}{y})= 3ln(y)-\frac{2y}{x}\]
\[y'\ = \frac{3ln(y)-\frac{2y}{x}}{2ln(x) -\frac{3x}{y}}\]

Answer 19

the answers ar different.... how do you get rid of the fractions?

Answer 20

algebra thru it ....

Answer 21

cus thats what i got

Answer 22

multiply top and bottom by xy

Answer 23

\[y'\ = \frac{3ln(y)-\frac{2y}{x}}{2ln(x) -\frac{3x}{y}}*\frac{xy}{xy}\]
\[y'\ = \frac{3xyln(y)-2yy}{2xyln(x) -3xx}\]
\[y'\ = \frac{3xy\ ln(y)-2y^2}{2xy\ ln(x) -3x^2}\]
maybe?

Answer 24

it might help to know what the answer they say should look like

Answer 25

cus thats what i got

Answer 26

1) (y^2(lnx-1))/(2ylnx-3x)
2) (2y^2(lnx+1))/(2ylnx-3x)
3) (y(3xlny-2y))/(x(2ylnx-3x))
4) (2y^2(lnx-1))/(2ylnx+3x)
5) (y(2x^2(lny)-3y))/(x(3y^2(lnx)-2x))

Answer 27

\[\frac{y^2(lnx-1)}{2y\ lnx-3x}\]
\[\frac{ 2y^2(lnx+1)}{(2y\ lnx-3x}\]
\[ \frac{y(3xlny-2y)}{x(2ylnx-3x)}=\frac{3xy\ lny-2y^2}{2xy\ lnx-3x^2}\]

Answer 28

id go 3