Question

For dy/dx of cos xy+x^3=y^3 so far I've got:
-y sinx+cos x dy/dx+3x^2-3y^2dy/dx
Is this right so far or not?

Answer 1

cos(xy) + x^3 = y^3

-sin(xy) (x'y+xy') + 3x^2 x' = 3y^2 y'

-sin(xy) (y+xy') + 3x^2 = 3y^2 y'

Answer 2

assuming ive read your inital setup correctly of course

Answer 3

Yeah that looks more like the available answers, but I'm not sure why it's that way.
Why is it cos (xy) and not cosx y?

Answer 4

let -sin(xy) = A

A(y+xy') + 3x^2 = 3y^2 y'

Ay + Axy' + 3x^2 = 3y^2 y'

Ay + 3x^2 = 3y^2 y' - Axy'

Ay + 3x^2 = (3y^2 - Ax) y'


Ay + 3x^2
---------- = y'
3y^2 - Ax

Answer 5

cos(xy) is what im assuming you meant with: cos xy

Answer 6

grouping symbols would help out alot in clearing these things up

Answer 7

\[\cos xy+x ^{3}=y ^{3}\]
diff.w.r.t x,
\[-\sin xy \left( x \frac{ dy }{dx }+y*1 \right)=3y ^{2}\frac{ dy }{ dx }\]
\[or -x \sin xy \frac{ dy }{ dx }-y \sin xy=3y ^{2}\frac{ dy }{ dx }\]
\[\left( 3y ^{2}+x \sin xy \right)\frac{ dy }{dx }=-y \sin xy\]
you can find dy/dx

Answer 8

@surjithayer I think you left out the 3x^2 but I got it anyways, and the rest of it matches what I've got.

Thanks for helping!

Answer 9

yw