Question

differentiate implicitly to find \\( \frac { d y } { d x } \\). then find the slope of the curve at the given point
\\( x ^ { 2 } y - 2 x ^ { 2 } - 4 = 0 ; \quad ( 2,3 ) \\)
\\( \frac { d y } { d x } = \square \\)

Explanation:

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

Differentiate \(x^{2}y-2x^{2}-4 = 0\) 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 \(-2x^{2}\) is \(-4x\), and the derivative of \(-4\) is \(0\).
So, \(\frac{d}{dx}(x^{2}y)-\frac{d}{dx}(2x^{2})-\frac{d}{dx}(4)=\frac{d}{dx}(0)\) gives \(2xy + x^{2}\frac{dy}{dx}-4x=0\).

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

Isolate the terms with \(\frac{dy}{dx}\):
\(x^{2}\frac{dy}{dx}=4x - 2xy\).
Then \(\frac{dy}{dx}=\frac{4x - 2xy}{x^{2}}=\frac{4 - 2y}{x}\) (divide numerator and denominator by \(x\), assuming \(x
eq0\)).

Step3: Find the slope at the point \((2,3)\)

Substitute \(x = 2\) and \(y = 3\) into \(\frac{dy}{dx}=\frac{4 - 2y}{x}\).
\(\frac{dy}{dx}\big|_{(x = 2,y = 3)}=\frac{4-2\times3}{2}=\frac{4 - 6}{2}=\frac{-2}{2}=-1\).

Answer:

\(\frac{dy}{dx}=\frac{4 - 2y}{x}\), and the slope at the point \((2,3)\) is \(-1\).