What is the implicit derivative of #1=x/y #?

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Answer 1 · 2018-05-04T11:12:38

#dy/dx=y/x#

Since #y=x#, #dy/dx=1#

We have #f(x,y)=x/y=1#

#x/y=xy^-1#

We first derivate with respect to #x# first:
#d/dx[xy^-1]=d/dx[1]#

#y^-1+xd/dx[y^-1]=0#

Using the chain rule, we get:
#d/dx=d/dy*dy/dx#

#y^-1+dy/dxxd/dx[y^-1]=0#

#y^-1+dy/dx-xy^-2=0#

#dy/dxxy^-2=y^-1#

#dy/dx=y^-1/(xy^-2)=y^2/(xy)=y/x#

Since, we know #y=x# we can say that #dy/dx=x/x=1#