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tan(xy)=x , find dy/dx
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implicit? It's been awhile
Implicit differentiation... hmm... let's find the derivative of both sides :)
yea implicit
yes. i think it is \[(xy' + y) (\sec^2(xy))=1\]
Yep
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now just solve for dy/dx
which is the same as y'
so \[xy' + y = 1 / \sec^2(xy)\] ?
Then subtract y from both sides then divide x from both sides
Ok. I don't know how to make it look like the answer choices.
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\[(1-ytan(xy)\sec(xy) )/ xtan(xy)\sec(xy)\] \[\frac{ \sec^2(xy)-y }{ x }\] \[\cos^2(xy)\] \[\frac{ \cos^2(xy)-y }{ x}\] \[\frac{ \cos^2(xy) }{ x }\]
It's the second last one
Because 1/sec(x) is cos(x)
ahh ok. thanks!
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