Question

differentiate implicitly to find $\frac{dy}{dx}$. then find the slope of the curve at the given point.
$x^{2}y - 3x^{3}-4 = 0$; $(2,7)$
$\frac{dy}{dx}=\square$
the slope of the curve at $(2,7)$ is $\square$. (simplify your answer.)

Explanation:

Step1: Differentiate each term with respect to \(x\)

Differentiate \(x^{2}y\) using the product rule \((uv)^\prime = u^\prime v+uv^\prime\) (where \(u = x^{2}\), \(v = y\)), \((x^{2}y)^\prime=2xy + x^{2}\frac{dy}{dx}\).
Differentiate \(-3x^{3}\) using the power rule \((x^{n})^\prime=nx^{n - 1}\), \((-3x^{3})^\prime=-9x^{2}\).
Differentiate \(-4\) (a constant) gives \(0\).
So, differentiating \(x^{2}y-3x^{3}-4 = 0\) with respect to \(x\) 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 \((2,7)\)

Substitute \(x = 2\) and \(y = 7\) into \(\frac{dy}{dx}=\frac{9x - 2y}{x}\).
\(\frac{dy}{dx}\big|_{x = 2,y = 7}=\frac{9\times2-2\times7}{2}=\frac{18 - 14}{2}\).

Answer:

\(\frac{dy}{dx}=\frac{9x - 2y}{x}\); The slope of the curve at \((2,7)\) is \(2\).