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DLS
 3 years ago
DE
DLS
 3 years ago
DE

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DLS
 3 years ago
Best ResponseYou've already chosen the best response.1\[xcosy+ycosx=\tan^{1}x^{2}\]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1\[\frac{d(xcosy)}{dx} + \frac{d(ycosx)}{dx} = d(\tan^{1} x^{2})\]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1\[cosyxsin \frac{dy}{dx} + \dfrac{(ycosx)}{dx}  ysinx=\frac{1}{1+x^{4}}\]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1\[\frac{dy}{dx}(cosxxsiny)=\frac{1}{1+x^{4}}cosy+ysinx\]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1\[\frac{dy}{dx}=\frac{1}{1+x^{4}(cosyxsiny})  \frac{cosy+ysinx}{cosxxsiny}\]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.12nd equation is wrong sorry

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0\(\large \frac{dy}{dx}=\frac{1}{(1+x^{4})(cosyxsiny)} \frac{cosyysinx}{cosxxsiny}\)

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1\[cosyxsiny \frac{dy}{dx} +cosx \frac{dy}{dx}ysinx=\frac{1}{1+x^{4}}\]

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0what u wrote now is correct.

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0u get this \(\large \frac{dy}{dx}=\frac{1}{(1+x^{4})(cosyxsiny)} \frac{cosyysinx}{cosxxsiny}\)

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1if u open ur bracket u wud still get cosy+ysinx i guess
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