Question

question 1 (1 point)
use implicit differentiation to find the slope of the line tangent to the curve 4x² + 2x+xy = 2 at the point (2, - 9).
-9
there is no tangent line at (2, - 9).
-2/0
-2
-9/2

Explanation:

Step1: Differentiate each term

Differentiate $4x^{2}+2x + xy=2$ with respect to $x$.
The derivative of $4x^{2}$ is $8x$ using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$. The derivative of $2x$ is $2$. For the term $xy$, use the product rule $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$, where $u = x$ and $v = y$. So $\frac{d}{dx}(xy)=y+x\frac{dy}{dx}$. The derivative of the constant $2$ is $0$.
We get $8x + 2+y+x\frac{dy}{dx}=0$.

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

Isolate $\frac{dy}{dx}$:
\[

$$\begin{align*} x\frac{dy}{dx}&=-8x - 2 - y\\ \frac{dy}{dx}&=\frac{-8x - 2 - y}{x} \end{align*}$$

\]

Step3: Substitute the point $(2,-9)$

Substitute $x = 2$ and $y=-9$ into $\frac{dy}{dx}$:
\[

$$\begin{align*} \frac{dy}{dx}&=\frac{-8\times2-2-(-9)}{2}\\ &=\frac{-16 - 2 + 9}{2}\\ &=\frac{-18 + 9}{2}\\ &=\frac{-9}{2} \end{align*}$$

\]

Answer:

$\frac{-9}{2}$