Question

find the slope of the tangent to the curve $y^{2}=\frac{x^{2}}{xy - 14}$ at $(6,3)$
$\frac{dy}{dx}=\square$
(enter undef if the slope is not defined at this point.)

Explanation:

Step1: Differentiate both sides with respect to \(x\)

Differentiate \(y^{2}=\frac{x^{2}}{xy - 14}\) using the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\) and chain - rule \((y^{n})^\prime=ny^{n - 1}\frac{dy}{dx}\).

Left - hand side: \(\frac{d}{dx}(y^{2}) = 2y\frac{dy}{dx}\)

Right - hand side: \(u = x^{2}\), \(u^\prime=2x\); \(v=xy - 14\), \(v^\prime=y + x\frac{dy}{dx}\)

\(\frac{d}{dx}(\frac{x^{2}}{xy - 14})=\frac{2x(xy - 14)-x^{2}(y + x\frac{dy}{dx})}{(xy - 14)^{2}}\)

So, \(2y\frac{dy}{dx}=\frac{2x(xy - 14)-x^{2}(y + x\frac{dy}{dx})}{(xy - 14)^{2}}\)

Step2: Substitute \(x = 6\) and \(y = 3\) into the equation

First, substitute \(x = 6\) and \(y = 3\) into \(xy-14\): \(6\times3-14 = 4\)

The equation becomes:

\(2\times3\frac{dy}{dx}=\frac{2\times6\times(6\times3 - 14)-6^{2}(3 + 6\frac{dy}{dx})}{4^{2}}\)

\(6\frac{dy}{dx}=\frac{12\times4-36(3 + 6\frac{dy}{dx})}{16}\)

Multiply both sides by \(16\):

\(96\frac{dy}{dx}=48-108 - 216\frac{dy}{dx}\)

\(96\frac{dy}{dx}+216\frac{dy}{dx}=48 - 108\)

\(312\frac{dy}{dx}=-60\)

\(\frac{dy}{dx}=-\frac{60}{312}=-\frac{5}{26}\)

Answer:

\(-\frac{5}{26}\)