find the slope of the tangent line to the curve (sqrt 2x+4y) + (sqrt 4xy) = 9.16

at the point (1,5)
dy/dx method implicit differantiation?

To find the slope of the tangent line to the given curve at the point (1,5) using implicit differentiation, we will follow these steps:

1. Differentiate both sides of the equation with respect to x.
d/dx[(√(2x + 4y)) + (√(4xy))] = d/dx[9.16]

2. Apply the Chain Rule. If y = f(x), then dy/dx = (dy/du) * (du/dx), where u is an intermediate variable.
d/dx[(√(2x + 4y)) + (√(4xy))] = d/(du)[√(2x + 4y)] * d/dx[(2x + 4y)] + d/(du)[√(4xy)] * d/dx[4xy] = 0

3. Solve for d/dx(y):
d/(du)[√(2x + 4y)] * d/dx[(2x + 4y)] + d/(du)[√(4xy)] * d/dx[4xy] = 0
(1/(2 * √(2x + 4y))) * (2 + 4 * dy/dx) + (1/(2 * √(4xy))) * (4y + 4x * dy/dx) = 0
(1/√(2x + 4y)) + 2 * dy/dx/√(2x + 4y) + (2y + 2x * dy/dx)/√(4xy) = 0

4. Solve for dy/dx:
(1/√(2x + 4y)) + 2 * dy/dx/√(2x + 4y) + (2y + 2x * dy/dx)/√(4xy) = 0
2 * dy/dx/√(2x + 4y) + (2y + 2x * dy/dx)/√(4xy) = -1/√(2x + 4y)
2 * dy/dx/√(2x + 4y) + (2y + 2x * dy/dx)/√(4xy) + 1/√(2x + 4y) = 0

5. Substitute the coordinates (1,5) into the obtained equation.
2 * dy/dx/√(2(1) + 4(5)) + (2(5) + 2(1) * dy/dx)/√(4(1)(5)) + 1/√(2(1) + 4(5)) = 0

6. Solve for dy/dx, which will be the slope of the tangent line at (1,5):
2 * dy/dx/√(12) + (10 + 2 * dy/dx)/√(20) + 1/√(12) = 0
2 * dy/dx/√(12) + (10 + 2 * dy/dx)/√(20) = -1/√(12)
(2 * dy/dx * √(20) + (10 + 2 * dy/dx) * √(12))/√(12 * 20) = -1/√(12)

7. Simplify the equation:
(2 * dy/dx * √(20) + 10 * √(12) + 2 * dy/dx * √(12))/√(240) = -1/√(12)
2 * dy/dx * √(20) + 10 * √(12) + 2 * dy/dx * √(12) = -√(240)/√(12)

8. Further simplify:
2 * dy/dx * (√(20) + √(12)) = -√(240) - 10 * √(12)
dy/dx = (-√(240) - 10 * √(12))/(2 * (√(20) + √(12)))

Thus, the slope of the tangent line to the curve (sqrt(2x + 4y)) + (sqrt(4xy)) = 9.16 at the point (1,5) is given by dy/dx = (-√(240) - 10 * √(12))/(2 * (√(20) + √(12))).

Yes, implicit differentation.

1/2(sqrt2x+4y) * (2 dx+4dy)= 1/2sqrt(4xy)* (4ydx+4xdy)=0

and solve for dy/dx. Have fun.