If $-y^{3}-y+2+y^{2}=x^{2}$ then find $\frac{d y}{d x}$ at the point $(4,-2)$ . $\newline$ Answer: $\left.\frac{d y}{d x}\right|_{(4,-2)}=$

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Grade 8

Evaluate a nonlinear function

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Q. If

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

then find

\frac{d y}{d x}

at the point

(4,-2)

\newline

Answer:

\left.\frac{d y}{d x}\right|_{(4,-2)}=

Differentiate with respect to $x$ : First, we need to differentiate both sides of the equation with respect to $x$ to find $\frac{dy}{dx}$ . $\newline$ Given equation: $-y^{3} - y + 2 + y^{2} = x^{2}$ $\newline$ Differentiating both sides with respect to $x$ , we get: $\newline$ $\frac{d}{dx}(-y^{3} - y + 2 + y^{2}) = \frac{d}{dx}(x^{2})$ $\newline$ Since $y$ is a function of $x$ , we will use the chain rule for differentiating terms involving $y$ .
Left side differentiation: Differentiating the left side of the equation with respect to $x$ :
$\frac{d}{dx}(-y^{3}) = -3y^{2} \cdot \frac{dy}{dx}$
$\frac{d}{dx}(-y) = -1 \cdot \frac{dy}{dx}$
$\frac{d}{dx}(2) = 0$ (since the derivative of a constant is $0$ )
$\frac{d}{dx}(y^{2}) = 2y \cdot \frac{dy}{dx}$
Adding these up, we get:
$-3y^{2} \cdot \frac{dy}{dx} - \frac{dy}{dx} + 0 + 2y \cdot \frac{dy}{dx}$
Simplifying, we have:
$(-3y^{2} - 1 + 2y) \cdot \frac{dy}{dx}$
Right side differentiation: Differentiating the right side of the equation with respect to $x$ : $\frac{d}{dx}(x^{2}) = 2x$
Solve for $\frac{dy}{dx}$ : Now we equate the derivatives from both sides: $\newline$ $(-3y^{2} - 1 + 2y) \cdot \frac{dy}{dx} = 2x$ $\newline$ To find $\frac{dy}{dx}$ , we need to solve for it: $\newline$ $\frac{dy}{dx} = \frac{2x}{-3y^{2} - 1 + 2y}$
Substitute point $(4,-2)$ : We substitute the point $(4,-2)$ into the equation to find the value of $\frac{dy}{dx}$ at that point: $\newline$ $\left(\frac{dy}{dx}\right)|_{(4,-2)} = \frac{2\cdot 4}{-3\cdot (-2)^{2} - 1 + 2\cdot (-2)}$
Perform calculations: Now we perform the calculations: $\newline$ $(\frac{dy}{dx})|_{(4,-2)} = \frac{8}{(-3\cdot 4 - 1 - 4)}$ $\newline$ $(\frac{dy}{dx})|_{(4,-2)} = \frac{8}{(-12 - 1 - 4)}$ $\newline$ $(\frac{dy}{dx})|_{(4,-2)} = \frac{8}{(-17)}$
Final answer: Simplifying the fraction, we get the final answer: $(\frac{dy}{dx})|_{(4,-2)} = \frac{-8}{17}$