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 } = \\)

the slope of the curve at \\( ( 2,3 ) \\) is \\( \square \\). (simplify your answer.)

Explanation:

Step1: Differentiate both sides of the equation

Differentiate \(x^{2}y - 2x^{2}-4 = 0\) with respect to \(x\) using the product rule \((uv)^\prime=u^\prime v + uv^\prime\) (where \(u = x^{2}\), \(v=y\)).
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 we have:
\(2xy+x^{2}\frac{dy}{dx}-4x = 0\)

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

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

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

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

Answer:

\(\frac{dy}{dx}=\frac{4 - 2y}{x}\)
The slope of the curve at \((2,3)\) is \(-1\)