Question

differentiate implicitly to find \\( \frac { d y } { d x } \\). then find the slope of the curve at the given point
\\( x ^ { 2 } y - 3 x ^ { 3 } - 8 = 0 ; \quad ( 2,8 ) \\)
\\( \frac { d y } { d x } = \square \\)
the slope of the curve at \\( ( 2,8 ) \\) is \\( \square \\). (simplify your answer.)

Explanation:

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

Differentiate \(x^{2}y-3x^{3}-8 = 0\) term - by - term.
Using the product rule \((uv)^\prime=u^\prime v + uv^\prime\) (where \(u = x^{2}\) and \(v=y\)), the derivative of \(x^{2}y\) is \(2xy+x^{2}\frac{dy}{dx}\).
The derivative of \(-3x^{3}\) is \(-9x^{2}\), and the derivative of \(-8\) is \(0\).
So, \(\frac{d}{dx}(x^{2}y-3x^{3}-8)=\frac{d}{dx}(0)\) gives \(2xy + x^{2}\frac{dy}{dx}-9x^{2}=0\).

Step2: Solve for \(\frac{dy}{dx}\)

Isolate the terms with \(\frac{dy}{dx}\):
\(x^{2}\frac{dy}{dx}=9x^{2}-2xy\).
Then \(\frac{dy}{dx}=\frac{9x^{2}-2xy}{x^{2}}=\frac{9x - 2y}{x}\) (\(x
eq0\)).

Step3: Find the slope at the point \((2,8)\)

Substitute \(x = 2\) and \(y = 8\) into \(\frac{dy}{dx}\).
\(\frac{dy}{dx}\mid_{x = 2,y = 8}=\frac{9\times2-2\times8}{2}\).
First, calculate the numerator: \(9\times2-2\times8=18 - 16=2\).
Then \(\frac{2}{2}=1\).

Answer:

\(\frac{dy}{dx}=\frac{9x - 2y}{x}\); The slope of the curve at \((2,8)\) is \(1\).