Question

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

Answer 1

The slope is `43`.

Explanation:

Given that `x^3-(xy)/(x-y)= C` and that `(1,4)` lies on the curve, we can find `C = 1-(4)/(-3)=7/3`.

So the curve we are being asked about has equation:

`x^3-(xy)/(x-y)= 7/3`, which is equivalent to

`3x^3(x-y) - 3(xy) = 7(x-y)`, for `x != y`.

Simplifying gets us the equation:

`3x^4-3x^3y-3xy = 7x-7y`.

Note that this equation is linear in `y`, so we could solve for `y` explicitly, but it's probably simpler to differentiate now. Then we can substitute the point and finish by solving for `dy/dx`.
(If you're serious about learning mathematics, try solving for `y` first and find out which way seems simpler to you.)

Differentiating implicitly gets us:

`12x^3-9x^2y-3x^3dy/dx-3y-3xdy/dx=7-7dy/dx`

Rather than solving algebraically first, I think it is simpler to substitute the values now.
(Again, trying the other order --solving for `y` before putting in values -- is a good exercise in algebra and in comparing methods of solution.)

At `(1,4)`, we get

`12-9(4)-3dy/dx-3(4)-3dy/dx=7-7dy/dx`

Do the artihmetic/algebra to get `dy/dx = 43`