How do you implicitly differentiate #-1=sin(x/y) #?

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Answer 1 · 2016-04-08T16:40:21

#y'=y/x#

#0=cos(x/y)[(y-xy')/y^2]#->Quotient Rule

#0=(ycos(x/y)-xy'cos(x/y))/y^2#

#0*y^2=ycos(x/y)-xy'cos(x/y)#

#0=ycos(x/y)-xy'cos(x/y)#

#xy'cos(x/y)=ycos(x/y)#

#y'=(ycos(x/y))/(xcos(x/y))#

#y'=y/x#