How do you use implicit differentiation to find #y'# for #sin(xy) = 1#?

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Answer 1 · 2014-09-12T23:42:30

By implicit differentiation, we can find
#y'=1/{xcos(xy)}-y/x#.

Let us work through it.

#sin(xy)=1#

by implicitly differentiating with respect to #x#,
#Rightarrow cos(xy)cdot(1cdot y+x cdot y')=1#

by dividing by #cos(xy)#,
#Rightarrow y+xy'=1/cos(xy)#

by subtracting #y#,
#Rightarrow xy'=1/cos(xy)-y#

by dividing by #x#,
#Rightarrow y'=1/{xcos(xy)}-y/x#