Question

find the slope of the tangent line to the curve

$2x^{2}-4xy - 2y^{3}=104$

at the point $(-2,-4)$.

hint: you will need to use the product rule on the middle term.

Explanation:

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

Differentiate \(2x^{2}-4xy - 2y^{3}\) term - by - term.
Using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\), \(\frac{d}{dx}(2x^{2})=4x\).
For the term \(-4xy\), use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u=-4x\) and \(v = y\). So \(\frac{d}{dx}(-4xy)=-4y-4x\frac{dy}{dx}\).
For the term \(-2y^{3}\), use the chain rule \(\frac{d}{dx}(f(y))=f^\prime(y)\frac{dy}{dx}\). So \(\frac{d}{dx}(-2y^{3})=-6y^{2}\frac{dy}{dx}\).
Differentiating the right - hand side \(\frac{d}{dx}(104) = 0\).
The derivative of the entire equation is:
\(4x-4y-4x\frac{dy}{dx}-6y^{2}\frac{dy}{dx}=0\)

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

Group the terms with \(\frac{dy}{dx}\) together:
\(-4x\frac{dy}{dx}-6y^{2}\frac{dy}{dx}=4y - 4x\)
Factor out \(\frac{dy}{dx}\):
\(\frac{dy}{dx}(-4x - 6y^{2})=4y - 4x\)
Then \(\frac{dy}{dx}=\frac{4y - 4x}{-4x - 6y^{2}}=\frac{2y - 2x}{-2x - 3y^{2}}\)

Step3: Substitute \(x=-2\) and \(y = - 4\) into \(\frac{dy}{dx}\)

Substitute \(x=-2\) and \(y=-4\) into \(\frac{dy}{dx}=\frac{2y - 2x}{-2x - 3y^{2}}\)
\(\frac{dy}{dx}=\frac{2(-4)-2(-2)}{-2(-2)-3(-4)^{2}}\)
First, calculate the numerator: \(2(-4)-2(-2)=-8 + 4=-4\)
Then, calculate the denominator: \(-2(-2)-3(-4)^{2}=4-3\times16=4 - 48=-44\)

Answer:

\(\frac{1}{11}\)