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If
-3-3x^(2)-5y=-y^(3)-y^(2) then find
(dy)/(dx) at the point
(-1,-2).
Answer:
(dy)/(dx)|_((-1,-2))=

If $-3-3 x^{2}-5 y=-y^{3}-y^{2}$ then find $\frac{d y}{d x}$ at the point $(-1,-2)$ . $\newline$ Answer: $\left.\frac{d y}{d x}\right|_{(-1,-2)}=$

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Evaluate a nonlinear function

Full solution

Q. If

-3-3 x^{2}-5 y=-y^{3}-y^{2}

then find

\frac{d y}{d x}

at the point

(-1,-2)

.

\newline

Answer:

\left.\frac{d y}{d x}\right|_{(-1,-2)}=

Rewrite equation: First, we need to rewrite the given equation to make it easier to differentiate with respect to $x$ . The given equation is: $-3 - 3x^2 - 5y = -y^3 - y^2$ We can rewrite this as: $3x^2 + 5y + 3 = y^3 + y^2$
Differentiate with respect to $x$ : Next, we differentiate both sides of the equation with respect to $x$ , keeping in mind that $y$ is a function of $x$ , so we will use the chain rule for terms involving $y$ .
$\frac{d}{dx}(3x^2 + 5y + 3) = \frac{d}{dx}(y^3 + y^2)$
This gives us: $6x + 5\frac{dy}{dx} = 3y^2\frac{dy}{dx} + 2y\frac{dy}{dx}$
Solve for $\frac{dy}{dx}$ : Now, we need to solve for $\frac{dy}{dx}$ . To do this, we collect all the terms involving $\frac{dy}{dx}$ on one side of the equation. $\newline$ $5\left(\frac{dy}{dx}\right) - 3y^2\left(\frac{dy}{dx}\right) - 2y\left(\frac{dy}{dx}\right) = -6x$ $\newline$ This simplifies to: $(5 - 3y^2 - 2y)\left(\frac{dy}{dx}\right) = -6x$
Isolate $\frac{dy}{dx}$ : We can now isolate $\frac{dy}{dx}$ by dividing both sides of the equation by $(5 - 3y^2 - 2y)$ . $\frac{dy}{dx} = \frac{-6x}{(5 - 3y^2 - 2y)}$
Substitute point: Finally, we substitute the point $(-1, -2)$ into the equation to find the value of $\frac{dy}{dx}$ at that point. $\newline$ $\left(\frac{dy}{dx}\right)|_{(-1,-2)} = \frac{-6(-1)}{(5 - 3(-2)^2 - 2(-2))}$ $\newline$ This simplifies to: $\left(\frac{dy}{dx}\right)|_{(-1,-2)} = \frac{6}{(5 - 3(4) + 4)}$
Calculate denominator: Now we calculate the denominator: $5 - 3(4) + 4 = 5 - 12 + 4 = -7 + 4 = -3$ $\newline$ So, $(dy/dx)|_{(-1,-2)} = 6 / -3$
Find $\frac{dy}{dx}$ : Finally, we find the value of $\frac{dy}{dx}$ at the point $(-1, -2)$ . $\left(\frac{dy}{dx}\right)|_{(-1,-2)} = \frac{6}{-3} = -2$

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