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

Differentiate Equation: Differentiate both sides of the equation with respect to x. We have y - x^3 + 2y^2 = 5. To find dy/dx, we need to differentiate each term with respect to x. Remember that y is a function of x, so we will use the chain rule for the terms involving y. Differentiating y with respect to x gives dy/dx. Differentiating -x^3 with respect to x gives -3x^2. Differentiating 2y^2 with respect to x gives 4y(dy/dx) using the chain rule. The constant 5 differentiates to 0. So, the differentiated equation is: dy/dx - 3x^2 + 4y(dy/dx) = 0Solve for dy/dx: Solve for dy/dx. We have dy/dx - 3x^2 + 4y(dy/dx) = 0. Rearrange the terms to isolate dy/dx: dy/dx + 4y(dy/dx) = 3x^2 Factor out dy/dx: dy/dx(1 + 4y) = 3x^2 Divide both sides by (1 + 4y) to solve for dy/dx: dy/dx = 3x^2 / (1 + 4y)Substitute Point: Substitute the point (1, -2) into the derivative to find the slope at that point. We have dy/dx = 3x^2 / (1 + 4y). Substitute x = 1 and y = -2: dy/dx|_((1,-2)) = 3(1)^2 / (1 + 4(-2)) Calculate the denominator: 1 + 4(-2) = 1 - 8 = -7 Calculate the numerator: 3(1)^2 = 3 Now, divide the numerator by the denominator: dy/dx|_((1,-2)) = 3 / (-7)

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

If $y-x^{3}+2 y^{2}=5$ 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-x^{3}+2 y^{2}=5

then find

\frac{d y}{d x}

at the point

(1,-2)

\newline

Answer:

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

Differentiate Equation: Differentiate both sides of the equation with respect to $x$ . We have $y - x^3 + 2y^2 = 5$ . To find $\frac{dy}{dx}$ , we need to differentiate each term with respect to $x$ . Remember that $y$ is a function of $x$ , so we will use the chain rule for the terms involving $y$ . Differentiating $y$ with respect to $x$ gives $\frac{dy}{dx}$ . Differentiating $y - x^3 + 2y^2 = 5$ $0$ with respect to $x$ gives $y - x^3 + 2y^2 = 5$ $2$ . Differentiating $y - x^3 + 2y^2 = 5$ $3$ with respect to $x$ gives $y - x^3 + 2y^2 = 5$ $5$ using the chain rule. The constant $y - x^3 + 2y^2 = 5$ $6$ differentiates to $y - x^3 + 2y^2 = 5$ $7$ . So, the differentiated equation is: $y - x^3 + 2y^2 = 5$ $8$
Solve for $\frac{dy}{dx}$ : Solve for $\frac{dy}{dx}$ . We have $\frac{dy}{dx} - 3x^2 + 4y\frac{dy}{dx} = 0$ . Rearrange the terms to isolate $\frac{dy}{dx}$ : $\frac{dy}{dx} + 4y\frac{dy}{dx} = 3x^2$ Factor out $\frac{dy}{dx}$ : $\frac{dy}{dx}(1 + 4y) = 3x^2$ Divide both sides by $(1 + 4y)$ to solve for $\frac{dy}{dx}$ : $\frac{dy}{dx} = \frac{3x^2}{(1 + 4y)}$
Substitute Point: Substitute the point $(1, -2)$ into the derivative to find the slope at that point. $\newline$ We have $\frac{dy}{dx} = \frac{3x^2}{1 + 4y}$ . $\newline$ Substitute $x = 1$ and $y = -2$ : $\newline$ $\frac{dy}{dx}\bigg|_{(1,-2)} = \frac{3(1)^2}{1 + 4(-2)}$ $\newline$ Calculate the denominator: $\newline$ $1 + 4(-2) = 1 - 8 = -7$ $\newline$ Calculate the numerator: $\newline$ $3(1)^2 = 3$ $\newline$ Now, divide the numerator by the denominator: $\newline$ $\frac{dy}{dx}\bigg|_{(1,-2)} = \frac{3}{-7}$