DLS
  • DLS
DE
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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DLS
  • DLS
\[xcosy+ycosx=\tan^{-1}x^{2}\]
DLS
  • DLS
this is how i did it
DLS
  • DLS
\[\frac{d(xcosy)}{dx} + \frac{d(ycosx)}{dx} = d(\tan^{-1} x^{2})\]

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More answers

DLS
  • DLS
\[cosy-xsin \frac{dy}{dx} + \dfrac{(ycosx)}{dx} - ysinx=\frac{1}{1+x^{4}}\]
DLS
  • DLS
\[\frac{dy}{dx}(cosx-xsiny)=\frac{1}{1+x^{4}}-cosy+ysinx\]
DLS
  • DLS
\[\frac{dy}{dx}=\frac{1}{1+x^{4}(cosy-xsiny}) - \frac{cosy+ysinx}{cosx-xsiny}\]
DLS
  • DLS
2nd equation is wrong sorry
hartnn
  • hartnn
\(\large \frac{dy}{dx}=\frac{1}{(1+x^{4})(cosy-xsiny)}- \frac{cosy-ysinx}{cosx-xsiny}\)
DLS
  • DLS
\[cosy-xsiny \frac{dy}{dx} +cosx \frac{dy}{dx}-ysinx=\frac{1}{1+x^{4}}\]
DLS
  • DLS
oh so mine is correct? :O
hartnn
  • hartnn
what u wrote now is correct.
DLS
  • DLS
finalans is correct right
hartnn
  • hartnn
u get this \(\large \frac{dy}{dx}=\frac{1}{(1+x^{4})(cosy-xsiny)}- \frac{cosy-ysinx}{cosx-xsiny}\)
hartnn
  • hartnn
cos y - y sin x
DLS
  • DLS
i got that only
hartnn
  • hartnn
u got cos y+ y sin x
DLS
  • DLS
ah minor mistakes :p
DLS
  • DLS
if u open ur bracket u wud still get -cosy+ysinx i guess

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