If -4=x^(2)+y^(3)-3x then find (dy)/(dx) at the point (4,-2). Answer: (dy)/(dx)|_((4,-2))=

Implicit Differentiation: To find the derivative dy/dx, we need to implicitly differentiate the given equation with respect to x. Given equation: -4 = x^2 + y^3 - 3x Differentiate both sides with respect to x. d/dx(-4) = d/dx(x^2) + d/dx(y^3) - d/dx(3x) Since the derivative of a constant is 0, d/dx(-4) = 0. The derivative of x^2 with respect to x is 2x. For y^3, since y is a function of x, we use the chain rule: d/dx(y^3) = 3y^2 * dy/dx. The derivative of -3x with respect to x is -3. Putting it all together, we get: 0 = 2x + 3y^2 * (dy/dx) - 3.Solving for dy/dx: Now we need to solve for dy/dx. 0 = 2x + 3y^2 * (dy/dx) - 3 Rearrange the equation to isolate dy/dx on one side: 3y^2 * (dy/dx) = 3 - 2x Divide both sides by 3y^2 to solve for dy/dx: (dy/dx) = (3 - 2x) / (3y^2)Substitution and Calculation: Next, we substitute the point (4, -2) into the equation for dy/dx. x = 4 and y = -2 (dy/dx) = (3 - 2(4)) / (3(-2)^2) Calculate the values: (dy/dx) = (3 - 8) / (3 * 4) (dy/dx) = (-5) / (12)

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If
-4=x^(2)+y^(3)-3x then find
(dy)/(dx) at the point
(4,-2).
Answer:
(dy)/(dx)|_((4,-2))=

If $-4=x^{2}+y^{3}-3 x$ 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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Algebra 1

Transformations of quadratic functions

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Q. If

-4=x^{2}+y^{3}-3 x

then find

\frac{d y}{d x}

at the point

(4,-2)

\newline

Answer:

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

Implicit Differentiation: To find the derivative $\frac{dy}{dx}$ , we need to implicitly differentiate the given equation with respect to $x$ . Given equation: $-4 = x^2 + y^3 - 3x$ Differentiate both sides with respect to $x$ . $\frac{d}{dx}(-4) = \frac{d}{dx}(x^2) + \frac{d}{dx}(y^3) - \frac{d}{dx}(3x)$ Since the derivative of a constant is $0$ , $\frac{d}{dx}(-4) = 0$ . The derivative of $x^2$ with respect to $x$ is $2x$ . For $x$ $0$ , since $x$ $1$ is a function of $x$ , we use the chain rule: $x$ $3$ . The derivative of $x$ $4$ with respect to $x$ is $x$ $6$ . Putting it all together, we get: $x$ $7$ .
Solving for dy/dx: Now we need to solve for $\frac{dy}{dx}$ . $0 = 2x + 3y^2 \cdot \left(\frac{dy}{dx}\right) - 3$ Rearrange the equation to isolate $\frac{dy}{dx}$ on one side: $3y^2 \cdot \left(\frac{dy}{dx}\right) = 3 - 2x$ Divide both sides by $3y^2$ to solve for $\frac{dy}{dx}$ : $\frac{dy}{dx} = \frac{3 - 2x}{3y^2}$
Substitution and Calculation: Next, we substitute the point $(4, -2)$ into the equation for $\frac{dy}{dx}$ . $x = 4$ and $y = -2$ $\frac{dy}{dx} = \frac{(3 - 2(4))}{(3(-2)^2)}$ Calculate the values: $\frac{dy}{dx} = \frac{(3 - 8)}{(3 \cdot 4)}$ $\frac{dy}{dx} = \frac{(-5)}{(12)}$