How do you implicitly differentiate # sqrt(3x+3y) + sqrt(3xy) = 17.5#?

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Answer 1 · 2016-11-26T17:18:03

#sqrt(3x+3y)+sqrt(3xy) = 17.5#

#sqrt(3x+3y)+sqrt(3xy) = sqrt3 sqrt(x+y)+sqrt3 sqrt(xy)#, so

#sqrt(3x+3y)+sqrt(3xy) = 17.5# is equivalent to

#sqrt(x+y)+sqrt(xy) = 17.5/sqrt3#

Differentiate both sides with respect to #x#, using the chain rule.

#1/(2sqrt(x+y))(1+dy/dx) + 1/(2sqrt(xy))(y+x dy/dx) = 0#

#1/(2sqrt(x+y))+ 1/(2sqrt(x+y))dy/dx + y/(2sqrt(xy))+x/(2sqrt(xy)) dy/dx) = 0#

#((sqrt(xy)+xsqrt(x+y))/(sqrt(xy)sqrt(x+y)))dy/dx = (-sqrt(xy)-ysqrt(x+y))/(sqrt(xy)sqrt(x+y)#

#dy/dx = (-sqrt(xy)-ysqrt(x+y))/(sqrt(xy)+xsqrt(x+y))#