If -5y^(3)-xy-x=-2 then find (dy)/(dx) in terms of x and y. Answer: (dy)/(dx)=

Differentiate Equation: Differentiate both sides of the equation with respect to x. We will use the power rule for differentiating y^3, the product rule for differentiating xy, and the constant rule for differentiating constants. d/dx(-5y^3 - xy - x) = d/dx(-2)Apply Differentiation Rules: Apply the differentiation rules to each term. For -5y^3, since y is a function of x, we use the chain rule and get -15y^2(dy/dx). For -xy, we apply the product rule and get -y - x(dy/dx). For -x, the derivative is -1. For -2, the derivative is 0. So, -15y^2(dy/dx) - y - x(dy/dx) - 1 = 0Rearrange to Solve: Rearrange the terms to solve for dy/dx. Combine like terms and factor out dy/dx. dy/dx(-15y^2 - x) = y + 1Isolate dy/dx: Isolate dy/dx. Divide both sides by (-15y^2 - x) to solve for dy/dx. dy/dx = (y + 1) / (-15y^2 - x)

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If
-5y^(3)-xy-x=-2 then find
(dy)/(dx) in terms of
x and
y.
Answer:
(dy)/(dx)=

If $-5 y^{3}-x y-x=-2$ then find $\frac{d y}{d x}$ in terms of $x$ and $y$ . $\newline$ Answer: $\frac{d y}{d x}=$

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Algebra 1

Transformations of quadratic functions

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Q. If

-5 y^{3}-x y-x=-2

then find

\frac{d y}{d x}

in terms of

x

and

y

\newline

Answer:

\frac{d y}{d x}=

Differentiate Equation: Differentiate both sides of the equation with respect to $x$ . We will use the power rule for differentiating $y^3$ , the product rule for differentiating $xy$ , and the constant rule for differentiating constants. $\frac{d}{dx}(-5y^3 - xy - x) = \frac{d}{dx}(-2)$
Apply Differentiation Rules: Apply the differentiation rules to each term. $\newline$ For $-5y^3$ , since $y$ is a function of $x$ , we use the chain rule and get $-15y^2\frac{dy}{dx}$ . $\newline$ For $-xy$ , we apply the product rule and get $-y - x\frac{dy}{dx}$ . $\newline$ For $-x$ , the derivative is $-1$ . $\newline$ For $-2$ , the derivative is $0$ . $\newline$ So, $y$ $0$
Rearrange to Solve: Rearrange the terms to solve for $\frac{dy}{dx}$ . Combine like terms and factor out $\frac{dy}{dx}$ . $\frac{dy}{dx}(-15y^2 - x) = y + 1$
Isolate $\frac{dy}{dx}$ : Isolate $\frac{dy}{dx}$ . $\newline$ Divide both sides by $(-15y^2 - x)$ to solve for $\frac{dy}{dx}$ . $\newline$ $\frac{dy}{dx} = \frac{y + 1}{-15y^2 - x}$