Calculus question need help?

Hi

Recall that when you differentiate a function of y, you use implicit differentiation:

d/dx[f(y)] = f'(y)*dy/dx

Combining that with the product rule and chain rule, we have:

√(2x + 4y) + √(4xy) = 16.6

√(2x + 4y) + 2√(xy) = 16.6

d/dx[√(2x + 4y) + 2√(xy)] = d/dx[16.6]

(1/2)(2 + 4*dy/dx)/√(2x + 4y) + 2(1/2)(y + x*dy/dx)/√(xy) = 0

(1 + 2*dy/dx)/√(2x + 4y) + (y + x*dy/dx)/√(xy) = 0

1/√(2x + 4y) + 2*dy/dx/√(2x + 4y) + √(y/x) + √(x/y)*dy/dx = 0

2*dy/dx/√(2x + 4y) + √(x/y)*dy/dx = -1/√(2x + 4) - √(y/x)

dy/dx*[2/√(2x + 4y) + √(x/y)] = -1/√(2x + 4y) - √(y/x)

dy/dx = [-1/√(2x + 4y) - √(y/x)]/[2/√(2x + 4y) + √(x/y)]

Plug in x = 4 and y = 7 (because we're finding dy/dx at (4, 7)):

dy/dx = [-1/√(2*4 + 4*7) - √(7/4)]/[2/√(2*4 + 4*7) + √(4/7)]

dy/dx = (1/58)(-119 + 15√7)

dy/dx ≈ -1.367

the slope of the tangent is dy/dx, and when you take the derivative of a function with a y in it you use the chain rule where y=u and du/dx would then just be dy/dx. once you take the derivative you will have a formula with (dy/dx)'s in it, then just algebraically solve for dy/dx and plug in the x and y values