Question

DE

Answer 1

\[xcosy+ycosx=\tan^{-1}x^{2}\]

Answer 2

this is how i did it

Answer 3

\[\frac{d(xcosy)}{dx} + \frac{d(ycosx)}{dx} = d(\tan^{-1} x^{2})\]

Answer 4

\[cosy-xsin \frac{dy}{dx} + \dfrac{(ycosx)}{dx} - ysinx=\frac{1}{1+x^{4}}\]

Answer 5

\[\frac{dy}{dx}(cosx-xsiny)=\frac{1}{1+x^{4}}-cosy+ysinx\]

Answer 6

\[\frac{dy}{dx}=\frac{1}{1+x^{4}(cosy-xsiny}) - \frac{cosy+ysinx}{cosx-xsiny}\]

Answer 7

2nd equation is wrong sorry

Answer 8

\(\large \frac{dy}{dx}=\frac{1}{(1+x^{4})(cosy-xsiny)}- \frac{cosy-ysinx}{cosx-xsiny}\)

Answer 9

\[cosy-xsiny \frac{dy}{dx} +cosx \frac{dy}{dx}-ysinx=\frac{1}{1+x^{4}}\]

Answer 10

oh so mine is correct? :O

Answer 11

what u wrote now is correct.

Answer 12

finalans is correct right

Answer 13

u get this
\(\large \frac{dy}{dx}=\frac{1}{(1+x^{4})(cosy-xsiny)}- \frac{cosy-ysinx}{cosx-xsiny}\)

Answer 14

cos y - y sin x

Answer 15

i got that only

Answer 16

u got cos y+ y sin x

Answer 17

ah minor mistakes :p

Answer 18

if u open ur bracket u wud still get -cosy+ysinx i guess