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

Mar 24, 2016

The slope is $43$.

Explanation:

Given that ${x}^{3} - \frac{x y}{x - y} = C$ and that $\left(1 , 4\right)$ lies on the curve, we can find $C = 1 - \frac{4}{- 3} = \frac{7}{3}$.

${x}^{3} - \frac{x y}{x - y} = \frac{7}{3}$, which is equivalent to

$3 {x}^{3} \left(x - y\right) - 3 \left(x y\right) = 7 \left(x - y\right)$, for $x \ne y$.

Simplifying gets us the equation:

$3 {x}^{4} - 3 {x}^{3} y - 3 x y = 7 x - 7 y$.

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 $\frac{\mathrm{dy}}{\mathrm{dx}}$.
(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:

$12 {x}^{3} - 9 {x}^{2} y - 3 {x}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} - 3 y - 3 x \frac{\mathrm{dy}}{\mathrm{dx}} = 7 - 7 \frac{\mathrm{dy}}{\mathrm{dx}}$

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 $\left(1 , 4\right)$, we get

$12 - 9 \left(4\right) - 3 \frac{\mathrm{dy}}{\mathrm{dx}} - 3 \left(4\right) - 3 \frac{\mathrm{dy}}{\mathrm{dx}} = 7 - 7 \frac{\mathrm{dy}}{\mathrm{dx}}$

Do the artihmetic/algebra to get $\frac{\mathrm{dy}}{\mathrm{dx}} = 43$