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

Question

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

Samantha Adkison · Accepted Answer

#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))#

William Abdalla · Answer

undefined To implicitly differentiate the equation sqrt(3x+3y) + sqrt(3xy) = 17.5, follow these steps:

1. Differentiate each term of the equation with respect to both x and y.
2. Use the chain rule for differentiating the square root terms.
3. Solve for the derivatives.

The result of implicit differentiation for this equation with respect to x and y respectively is:

For $ \frac{dx}{dy} $:
$$ \frac{dx}{dy} = -\frac{3x + 3y}{3x + 6y + 3\sqrt{3x + 3y}} - \frac{3y}{3x + 6y + 3\sqrt{3xy}} $$

For $ \frac{dy}{dx} $:
$$ \frac{dy}{dx} = -\frac{3x}{3x + 6y + 3\sqrt{3x + 3y}} - \frac{3x + 3y}{3x + 6y + 3\sqrt{3xy}} $$