the questions in level 1 are introductory problems. the hints contain links to videos covering this content…

the questions in level 1 are introductory problems. the hints contain links to videos covering this content. question 1 (1 point) use implicit differentiation to find the slope of the line tangent to the curve 4x^{2}+2x + xy = 2 at the point (2, - 9). -2 there is no tangent line at (2, - 9). $\frac{-2}{9}$ $\frac{-9}{2}$ -9 view hint for question 1 question 2 (1 point) let f(x)=x^{6x}. then f(x) equals

the questions in level 1 are introductory problems. the hints contain links to videos covering this content. question 1 (1 point) use implicit differentiation to find the slope of the line tangent to the curve 4x^{2}+2x + xy = 2 at the point (2, - 9). -2 there is no tangent line at (2, - 9). $\frac{-2}{9}$ $\frac{-9}{2}$ -9 view hint for question 1 question 2 (1 point) let f(x)=x^{6x}. then f(x) equals

Answer

Explanation:

Step1: Differentiate each term

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

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

Isolate $\frac{dy}{dx}$: $x\frac{dy}{dx}=-8x - 2 - y$. $\frac{dy}{dx}=\frac{-8x - 2 - y}{x}$.

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

Substitute $x = 2$ and $y=-9$ into $\frac{dy}{dx}$: $\frac{dy}{dx}=\frac{-8\times2 - 2-(-9)}{2}=\frac{-16 - 2 + 9}{2}=\frac{-9}{2}$.

Answer:

$\frac{-9}{2}$