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.

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

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

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

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

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