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DLSBest ResponseYou've already chosen the best response.1
\[xcosy+ycosx=\tan^{1}x^{2}\]
 one year ago

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

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

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

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

DLSBest ResponseYou've already chosen the best response.1
2nd equation is wrong sorry
 one year ago

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

DLSBest ResponseYou've already chosen the best response.1
\[cosyxsiny \frac{dy}{dx} +cosx \frac{dy}{dx}ysinx=\frac{1}{1+x^{4}}\]
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
what u wrote now is correct.
 one year ago

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

DLSBest ResponseYou've already chosen the best response.1
if u open ur bracket u wud still get cosy+ysinx i guess
 one year ago
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