How do you differentiate #sin(xy)=1/2#?

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Answer 1 · 2017-01-10T19:56:57

Here are two ways to do it.

The easy (trick) way

From #sin(xy)=1/2# we conclude that #xy = pi/6 + 2pik# for integer #k#.

So #d/dx(xy) = d/dx(pi/6 + 2pik)#. That is,

#y+x dy/dx = 0#, and

#dy/dx = -y/x#

The non-easy way

#d/dx(sin(xy)) = d/dx(1/2)#

#cos(xy)(d/dx(xy)) = 0#

#cos(xy)(y + x dy/dx) = 0#

Divide by #cos(xy)# (OK, since we know it is not #0#.)

to get

#y+x dy/dx = 0#, and

#dy/dx = -y/x#