Find the slope of the tangent line to the curve below at a point

Question

Find the slope of the tangent line to the curve below at the point (7,5)

sqrt x+y + sqrt2xy = 11.830701880479

*sqrt is over both x+y*

slope=

Answer 1

By implicit derivation:

(1+y')/(2√(x+y) + (y+xy')/√(2xy) = 0

Setting x and y values

(1+y')/(2√12) +(5+7y')/√70) = 0

Solving for y'

slope, m = y' = -(10√12 + √70)/(√70+28√12) = -0.75632384486

If you plot the curve and y =m(x-7)+5 you will see that y is the tangent line at (7,5)

Answer 2

1/(2√x + y)·(1 + y') + 1/(2√2xy)(2y + 2xy') = 0. Point (7, 5) belong to this curve. Because

1/(2√12)(1 + y') + 1/(2√70)·(10 + 14y') = 0; (1 + y')/(4√3) + (5 + 7y')/√70 = 0;

1/(4√3) + y'/(4√3) + 5/√70 + 7y'/√70 = 0; y'(1/(4√3) + 7/√70) = - (1/(4√3) + 5/√70)

y' = - ( 1/(4√3) + 5/√70)/(1/(4√3) + 7/√70) = - 0.7563238449