What is the implicit derivative of #1= 3y-ysqrt(x-y) #?

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Answer 1 · 2018-03-26T00:46:48

#dy/dx = y/(3y-2x+6sqrt(x-y))#

You differentiate both sides of the expression

#1= 3y-ysqrt(x-y)#

with respect to #x# to get

#0 = 3 dy/dx- dy/dx sqrt{x-y}-y1/(2sqrt(x-y)) (1-dy/dx)#
#quad = (3-sqrt{x-y}+y/(2sqrt(x-y)))dy/dx-y/(2sqrt(x-y)) implies#

# (6sqrt(x-y)-2(x-y)+y)dy/dx-y=0 implies#

#(3y-2x+6sqrt(x-y))dy/dx = y implies#

#dy/dx = y/(3y-2x+6sqrt(x-y))#