find the slope of the tangent to the curve $y^{2}=\\frac{x^{2}}{xy - 14}$ at $(6,3)$\n$\\frac{dy}{dx}=\\squar…

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})