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find dy/dx by implicit differentiation
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\[e^ycosx=1+\sin(xy)\]
I am not sure how to differentiate sin(xy)
what is differentiation of sin(.) ??
cos(.)
is that right?
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Good..
Then you do the differentiation of .
\[\frac{d}{dx}(\sin(\color{red}{x})) = \cos(x) \times \frac{d}{dx} (\color{red}x)\]
\[\frac{d}{dx} (x) = 1\\ so, \; \; \frac{d}{dx}(\sin(x))= \cos(x) \times 1 = \cos(x)\]
yes
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Similarly do that..
\[\frac{d}{dx} (\sin(xy)) = \cos(xy) \times \frac{d}{dx}(xy)\]
oooh now i get it, so will it ultimately be cos x+ cos(xy)d/dx
Now, here product rule applies..
wait..
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ok
Final answer?
\[d(\sin(xy)) = \cos(xy)[x \frac{dy}{dx} + y]\]
Is this okay?
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