Question

find the slope of the tangent line to the curve -x^2 - 2xy - 3y^3 = 199 at the point (1, - 4). question help: video jump to answer

Explanation:

Step1: Differentiate implicitly

Differentiate $-x^{2}-2xy - 3y^{3}=199$ with respect to $x$.
Using the sum - rule and product - rule, we have:
$\frac{d}{dx}(-x^{2})+\frac{d}{dx}(-2xy)+\frac{d}{dx}(-3y^{3})=\frac{d}{dx}(199)$.
For $\frac{d}{dx}(-x^{2})=-2x$, for $\frac{d}{dx}(-2xy)$ using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u=-2x$ and $v = y$, we get $-2y-2x\frac{dy}{dx}$, and for $\frac{d}{dx}(-3y^{3})=-9y^{2}\frac{dy}{dx}$, and $\frac{d}{dx}(199) = 0$.
So, $-2x-2y - 2x\frac{dy}{dx}-9y^{2}\frac{dy}{dx}=0$.

Step2: Solve for $\frac{dy}{dx}$

Group the terms with $\frac{dy}{dx}$ on one side:
$-2x\frac{dy}{dx}-9y^{2}\frac{dy}{dx}=2x + 2y$.
Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(-2x-9y^{2})=2x + 2y$.
Then $\frac{dy}{dx}=\frac{2x + 2y}{-2x-9y^{2}}$.

Step3: Substitute the point $(1,-4)$

Substitute $x = 1$ and $y=-4$ into $\frac{dy}{dx}$:
$\frac{dy}{dx}=\frac{2(1)+2(-4)}{-2(1)-9(-4)^{2}}=\frac{2-8}{-2 - 144}=\frac{-6}{-146}=\frac{3}{73}$.

Answer:

$\frac{3}{73}$