How do you implicitly differentiate #-y=x-sqrt(x-y) #?

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Answer 1 · 2015-12-30T04:36:43

#dy/dx=(1-2sqrt(x-y))/(1+2sqrt(x-y))#

Find the derivative of each part.

#d/dx(-y)=-y'#

#d/dx(x)=1#

#d/dxsqrt(x-y)=1/(2sqrt(x-y))*(1-y')=(1-y')/(2sqrt(x-y))#

Combine them all (sum rule):

#-y'=1-(1-y')/(2sqrt(x-y))#

Multiply everything by #2sqrt(x-y)#.

#-2y'sqrt(x-y)=2sqrt(x-y)-1+y'#

Isolate #y'#.

#1-2sqrt(x-y)=y'+2y'sqrt(x-y)#

#y'(1+2sqrt(x-y))=1-2sqrt(x-y)#

#y'=(1-2sqrt(x-y))/(1+2sqrt(x-y))=dy/dx#