4x-x^(2)y+y^(3)=10 Find the value of (dy)/(dx) at the point (1,2). Choose 1 answer: (A) 0 (B) (2)/(3) (C) -1 (D) -4

Apply Rules and Derivatives: Now, apply the product rule to the term -x^2y and the chain rule to y^3. d/dx(4x) - d/dx(x^2y) + d/dx(y^3) = 0 4 - (2xy + x^2(dy/dx)) + 3y^2(dy/dx) = 0Substitute Point into Equation: Substitute the point (1,2) into the differentiated equation. 4 - (2*1*2 + 1^2(dy/dx)) + 3*2^2(dy/dx) = 0 4 - (4 + (dy/dx)) + 12(dy/dx) = 0Simplify and Solve for dy/dx: Simplify the equation and solve for dy/dx. 4 - 4 - dy/dx + 12dy/dx = 0 0 = dy/dx(12 - 1) dy/dx = 0/(11) dy/dx = 0

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4x-x^(2)y+y^(3)=10
Find the value of
(dy)/(dx) at the point
(1,2).
Choose 1 answer:
(A) 0
(B)
(2)/(3)
(C) -1
(D) -4

$4 x-x^{2} y+y^{3}=10$ $\newline$ Find the value of $\frac{d y}{d x}$ at the point $(1,2)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $0$ $\newline$ (B) $\frac{2}{3}$ $\newline$ (C) $-1$ $\newline$ (D) $-4$

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Math Problems

Algebra 1

Transformations of quadratic functions

Full solution

4 x-x^{2} y+y^{3}=10

\newline

Find the value of

\frac{d y}{d x}

at the point

(1,2)

\newline

Choose

1

answer:

\newline

(A)

0

\newline

(B)

\frac{2}{3}

\newline

(C)

-1

\newline

(D)

-4

Apply Rules and Derivatives: Now, apply the product rule to the term $-x^2y$ and the chain rule to $y^3$ . $\newline$ $\frac{d}{dx}(4x) - \frac{d}{dx}(x^2y) + \frac{d}{dx}(y^3) = 0$ $\newline$ $4 - (2xy + x^2\frac{dy}{dx}) + 3y^2\frac{dy}{dx} = 0$
Substitute Point into Equation: Substitute the point $(1,2)$ into the differentiated equation. $\newline$ $4 - (2\cdot 1\cdot 2 + 1^2\frac{dy}{dx}) + 3\cdot 2^2\frac{dy}{dx} = 0$ $\newline$ $4 - (4 + \frac{dy}{dx}) + 12\frac{dy}{dx} = 0$
Simplify and Solve for $\frac{dy}{dx}$ : Simplify the equation and solve for $\frac{dy}{dx}$ . $4 - 4 - \frac{dy}{dx} + 12\frac{dy}{dx} = 0$ $0 = \frac{dy}{dx}(12 - 1)$ $\frac{dy}{dx} = \frac{0}{11}$ $\frac{dy}{dx} = 0$