Question

What is the slope of the tangent line of `(1-x)(3-4y^2)-lny = C`, where C is an arbitrary constant, at `(1,1)`?

Answer 1

`m=1`

Explanation:

`f(x,y)=(1-x)(3-4y^2)-lny-C`

`dy/dx = - f_x/f_y`

`= -((-1)(3-4y^2))/((1-x)(-8y)-1/y)`

At `(1,1)`, we get

`dy/dx = -((-1)(-1))/((0)(-8)-1) = 1`