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

1 Answer
Jim H
Nov 26, 2016

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

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

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

sqrt(x+y)+sqrt(xy) = 17.5/sqrt3√x+y+√xy=17.5√3

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

1/(2sqrt(x+y))(1+dy/dx) + 1/(2sqrt(xy))(y+x dy/dx) = 012√x+y(1+dydx)+12√xy(y+xdydx)=0

1/(2sqrt(x+y))+ 1/(2sqrt(x+y))dy/dx + y/(2sqrt(xy))+x/(2sqrt(xy)) dy/dx) = 012√x+y+12√x+ydydx+y2√xy+x2√xydydx)=0

((sqrt(xy)+xsqrt(x+y))/(sqrt(xy)sqrt(x+y)))dy/dx = (-sqrt(xy)-ysqrt(x+y))/(sqrt(xy)sqrt(x+y)(√xy+x√x+y√xy√x+y)dydx=−√xy−y√x+y√xy√x+y

dy/dx = (-sqrt(xy)-ysqrt(x+y))/(sqrt(xy)+xsqrt(x+y))dydx=−√xy−y√x+y√xy+x√x+y

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