Question

find dy/dx if lnx-lny^2+sinxy=8

Answer 1

what is the derivative of ln(x) ?

Answer 2

and then of: -lny^2

and of sin(xy)

and lastly of 8

Answer 3

can i send the wolfram ans

Answer 4

for this question

Answer 5

why?

Answer 6

for clearifying

Answer 7

I had trouble with the last term, but found an answer here:
http://www.scienceforums.net/topic/23152-implicit-diff/

Answer 8

http://www.wolframalpha.com/input/?i=lnx-lny%5E2%2Bsinxy%3D8+find+dy%2Fdx+

Answer 9

the last term? really? 8' = 0

Answer 10

second to last :)

Answer 11

clarifying what? there wasnt even anything that needed to be clarified yet...

Answer 12

this question but i think the ans is wrong in wolfram

Answer 13

sin(xy); use chain rule

cos(xy) * (xy)' ; use product rule

cos(xy) * (x'y + xy') ; since dx/dx=x' = 1, ignore the x's

cos(xy)(y+xy')

Answer 14

@ParthKohli

Answer 15

usually askers appreciate being guided and taught rather than being given an answer @best.shakir :)

Answer 16

i also have doubt in this question

Answer 17

@amistre64 thank u

Answer 18

yalls welcome :)

Answer 19

just to add little bit here
this requires Implicit differentiation. and for the second term derivative should be
\[\Large \frac{d}{dx}\ln(y^2) =\frac{1}{y^2}*2y \frac{dy}{dx}\].