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

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this is how i did it

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

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

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

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\[\frac{dy}{dx}=\frac{1}{1+x^{4}(cosyxsiny})  \frac{cosy+ysinx}{cosxxsiny}\]

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2nd equation is wrong sorry

hartnn
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\(\large \frac{dy}{dx}=\frac{1}{(1+x^{4})(cosyxsiny)} \frac{cosyysinx}{cosxxsiny}\)

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

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oh so mine is correct? :O

hartnn
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what u wrote now is correct.

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finalans is correct right

hartnn
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u get this
\(\large \frac{dy}{dx}=\frac{1}{(1+x^{4})(cosyxsiny)} \frac{cosyysinx}{cosxxsiny}\)

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cos y  y sin x

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i got that only

hartnn
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u got cos y+ y sin x

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ah minor mistakes :p

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