If -y^(3)+2y-x^(3)-3=0 then find (dy)/(dx) at the point (1,-2). Answer: (dy)/(dx)|_((1,-2))=

Implicit Differentiation: To find the derivative of y with respect to x, (dy)/(dx), we need to implicitly differentiate the given equation with respect to x.Solving for (dy)/(dx): Differentiate both sides of the equation with respect to x. Remember that when differentiating y terms with respect to x, we treat y as a function of x and use the chain rule to include (dy)/(dx). Differentiating -y^3 with respect to x gives us -3y^2(dy)/(dx). Differentiating 2y with respect to x gives us 2(dy)/(dx). Differentiating -x^3 with respect to x gives us -3x^2. Differentiating -3 with respect to x gives us 0, since it's a constant. So, the differentiated equation is -3y^2(dy)/(dx) + 2(dy)/(dx) - 3x^2 = 0.Substitution and Calculation: Now, we solve for (dy)/(dx) by isolating it on one side of the equation. Group the terms containing (dy)/(dx) together: (-3y^2 + 2)(dy)/(dx) - 3x^2 = 0. Add 3x^2 to both sides: (-3y^2 + 2)(dy)/(dx) = 3x^2. Divide both sides by (-3y^2 + 2) to get (dy)/(dx) = 3x^2 / (-3y^2 + 2).Substitute the given point (1, -2) into the equation to find the value of (dy)/(dx) at that point. (dy)/(dx) = 3(1)^2 / (-3(-2)^2 + 2). (dy)/(dx) = 3 / (-3(4) + 2). (dy)/(dx) = 3 / (-12 + 2). (dy)/(dx) = 3 / (-10). (dy)/(dx) = -3/10.

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If
-y^(3)+2y-x^(3)-3=0 then find
(dy)/(dx) at the point
(1,-2).
Answer:
(dy)/(dx)|_((1,-2))=

If $-y^{3}+2 y-x^{3}-3=0$ 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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Algebra 1

Transformations of quadratic functions

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Q. If

-y^{3}+2 y-x^{3}-3=0

then find

\frac{d y}{d x}

at the point

(1,-2)

\newline

Answer:

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

Implicit Differentiation: To find the derivative of $y$ with respect to $x$ , $\frac{dy}{dx}$ , we need to implicitly differentiate the given equation with respect to $x$ .
Solving for $(\frac{dy}{dx}):</b> Differentiate both sides of the equation with respect to \$x$ . Remember that when differentiating $y$ terms with respect to $x$ , we treat $y$ as a function of $x$ and use the chain rule to include $(\frac{dy}{dx})$ . $\newline$ Differentiating $-y^3$ with respect to $x$ gives us $-3y^2(\frac{dy}{dx})$ . $\newline$ Differentiating $2y$ with respect to $x$ gives us $y$ $1$ . $\newline$ Differentiating $y$ $2$ with respect to $x$ gives us $y$ $4$ . $\newline$ Differentiating $y$ $5$ with respect to $x$ gives us $y$ $7$ , since it's a constant. $\newline$ So, the differentiated equation is $y$ $8$ .
Substitution and Calculation: Now, we solve for $(dy)/(dx)$ by isolating it on one side of the equation. $\newline$ Group the terms containing $(dy)/(dx)$ together: $(-3y^2 + 2)(dy)/(dx) - 3x^2 = 0$ . $\newline$ Add $3x^2$ to both sides: $(-3y^2 + 2)(dy)/(dx) = 3x^2$ . $\newline$ Divide both sides by $(-3y^2 + 2)$ to get $(dy)/(dx) = 3x^2 / (-3y^2 + 2)$ .
Substitution and Calculation: Now, we solve for $\frac{dy}{dx}$ by isolating it on one side of the equation. $\newline$ Group the terms containing $\frac{dy}{dx}$ together: $(-3y^2 + 2)\frac{dy}{dx} - 3x^2 = 0$ . $\newline$ Add $3x^2$ to both sides: $(-3y^2 + 2)\frac{dy}{dx} = 3x^2$ . $\newline$ Divide both sides by $(-3y^2 + 2)$ to get $\frac{dy}{dx} = \frac{3x^2}{-3y^2 + 2}$ .Substitute the given point $(1, -2)$ into the equation to find the value of $\frac{dy}{dx}$ at that point. $\newline$ $\frac{dy}{dx} = \frac{3(1)^2}{-3(-2)^2 + 2}$ . $\newline$ $\frac{dy}{dx}$ $0$ . $\newline$ $\frac{dy}{dx}$ $1$ . $\newline$ $\frac{dy}{dx}$ $2$ . $\newline$ $\frac{dy}{dx}$ $3$ .

If $-y^{3}+2 y-x^{3}-3=0$ 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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If $-y^{3}+2 y-x^{3}-3=0$ then find $\frac{d y}{d x}$ at the point $(1,-2)$ . $\newline$ Answer: $\left.\frac{d y}{d x}\right|_{(1,-2)}=$