e^(2x-y) = x^2 y^-1, tangent line the the graph (2,4)

Question

The equation to the tangent line to the graph?

Answer 1

To find the derivative, first take the natural log of both sides then use implicit differentiation:

ln(e^2x-y) = ln(x²/y)

2x - y = 2ln(x) - ln(y)

d/dx(2x - y) = d/dx(2ln(x) - ln(y))

2 - dy/dx = 2/x - 1/y(dy/dx)

2 -2/x = dy/dx - 1/y(dy/dx)

2x/x - 2/x = dy/dx(1 - 1/y)

(2x - 2)/x = dy/dx(y/y - 1/y)

2(x - 1)/x = dy/dx(y - 1)/y

dy/dx = 2y(x - 1)/(x(y - 1))

Plug in (2, 4) to get the value of the slope at (2, 4) then plug that value of "m" into the point-slope form: y - y1 = m(x - x1) where (x1, y1) = (2, 4)