Question

if e^xy−y^2=e−4, then at x = 1/2 and y = 2, dy/dx =

Answer 1

Answer:

dy/dx = -0.5285

Step-by-step explanation:

e^(xy) − y² = e^(−4)

To get dy/dx, we will differentiate with respect to x applying both chain rule and product rule.

Thus;

e^(xy)[d(xy)/dx] - (d/dx)y² = 0

e^(xy)[x(dy/dx) + y] - ((dy/dx) × (d(y²)/dy)) = 0

xe^(xy)[dy/dx] + ye^(xy) - 2y = 0

Thus, rearranging we have;

dy/dx = [2y -(ye^(xy))]/xe^(xy)

We are given;

x = 1/2 and y = 2

Thus;

dy/dx = [2(2) - (2e^(½ × 2))]/(1e^(½ × 2))

dy/dx = (4 - 5.4366)/2.7183

dy/dx = -0.5285