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anonymous
 3 years ago
xcos yy =cosxy
by implicit differentiation pls help!!
anonymous
 3 years ago
xcos yy =cosxy by implicit differentiation pls help!!

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AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0ok tell me where are you stuck in doing this question?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i dont have no idea in solving that...its our assignment and it was not been discussed

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0ok i suggest you differentiate both sides and bring y' to one side

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1@jayshane We have to differentiate both sides. Hmm... shall we differentiate term by term? I think it would be easier for you to understand..

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0keep in mind \[\large \bf \text{differential of sum =sum of differentials }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0example equation pls!!!

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1Example: Differentiate a sinb with respect to a \[\frac{d}{da}asinb=sinb\frac{d}{da}a + a\frac{d}{da}sinb =sinb + a(cosb)(b')\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sin xyy`=sin xyy` ?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1For cosxy \[\frac{d}{dx}cos(xy) = \frac{d}{du}cos(u) \times \frac{d}{dx}(xy)\]u= xy\[=sinu \times (y\frac{d}{dx}x +x\frac{d}{dx}x)\]\[=...\]
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