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jayshane
Group Title
xcos yy =cosxy
by implicit differentiation pls help!!
 one year ago
 one year ago
jayshane Group Title
xcos yy =cosxy by implicit differentiation pls help!!
 one year ago
 one year ago

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AravindG Group TitleBest ResponseYou've already chosen the best response.0
ok tell me where are you stuck in doing this question?
 one year ago

jayshane Group TitleBest ResponseYou've already chosen the best response.0
i dont have no idea in solving that...its our assignment and it was not been discussed
 one year ago

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

Callisto Group TitleBest 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..
 one year ago

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

jayshane Group TitleBest ResponseYou've already chosen the best response.0
example equation pls!!!
 one year ago

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

jayshane Group TitleBest ResponseYou've already chosen the best response.0
sin xyy`=sin xyy` ?
 one year ago

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