Question

if (3x^{3}+2x + xy=3) and (y(3)= - 10), find (y(3)) by implicit differentiation.

Explanation:

Step1: Differentiate both sides

Differentiate $3x^{3}+2x + xy=3$ with respect to $x$.
Using the sum - rule of differentiation, we have $\frac{d}{dx}(3x^{3})+\frac{d}{dx}(2x)+\frac{d}{dx}(xy)=\frac{d}{dx}(3)$.
For $\frac{d}{dx}(3x^{3})$, by the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we get $9x^{2}$.
For $\frac{d}{dx}(2x)$, we get 2.
For $\frac{d}{dx}(xy)$, by the product - rule $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$ where $u = x$ and $v = y$, we get $x\frac{dy}{dx}+y$.
And $\frac{d}{dx}(3)=0$.
So, $9x^{2}+2+x\frac{dy}{dx}+y = 0$.

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

Rearrange the equation $9x^{2}+2+x\frac{dy}{dx}+y = 0$ to isolate $\frac{dy}{dx}$.
We get $x\frac{dy}{dx}=-9x^{2}-y - 2$.
Then $\frac{dy}{dx}=\frac{-9x^{2}-y - 2}{x}$.

Step3: Substitute $x = 3$ and $y(3)=-10$

Substitute $x = 3$ and $y=-10$ into $\frac{dy}{dx}=\frac{-9x^{2}-y - 2}{x}$.
$\frac{dy}{dx}\big|_{x = 3}=\frac{-9\times3^{2}-(-10)-2}{3}$.
First, calculate the numerator: $-9\times3^{2}-(-10)-2=-9\times9 + 10-2=-81 + 10-2=-73$.
Then $\frac{dy}{dx}\big|_{x = 3}=\frac{-73}{3}$.

Answer:

$-\frac{73}{3}$