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

Evaluate an exponential function

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Q. If

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

then find

\frac{d y}{d x}

in terms of

x

and

y

\newline

Answer:

\frac{d y}{d x}=

Differentiate Left Side: We are given the equation $-2y^2 - x = 2y^3 + x^3$ . To find $\frac{dy}{dx}$ , we need to differentiate both sides of the equation with respect to $x$ , treating $y$ as a function of $x$ (implicit differentiation).
Differentiate Right Side: Differentiate the left side of the equation with respect to $x$ : $-2y^2 - x$ . The derivative of $-2y^2$ with respect to $x$ is $-4y\frac{dy}{dx}$ because $y$ is a function of $x$ . The derivative of $-x$ with respect to $x$ is $-1$ .
Combine Differentiated Equations: Differentiate the right side of the equation with respect to $x$ : $2y^3 + x^3$ . The derivative of $2y^3$ with respect to $x$ is $6y^2(\frac{dy}{dx})$ because $y$ is a function of $x$ . The derivative of $x^3$ with respect to $x$ is $3x^2$ .
Solve for $\frac{dy}{dx}$ : Now we have the differentiated equation: $-4y\frac{dy}{dx} - 1 = 6y^2\frac{dy}{dx} + 3x^2$ .
Move Terms to Respective Sides: We need to solve for $\frac{dy}{dx}$ . To do this, we'll move all the terms involving $\frac{dy}{dx}$ to one side of the equation and the rest to the other side. This gives us $-4y\frac{dy}{dx} - 6y^2\frac{dy}{dx} = 3x^2 + 1$ .
Combine Like Terms: Combine like terms involving $\frac{dy}{dx}$ on the left side: $-4y\frac{dy}{dx} - 6y^2\frac{dy}{dx}$ becomes $(-4y - 6y^2)\frac{dy}{dx}$ .
Isolate $(\frac{dy}{dx}):</b> Now we have \$(-4y - 6y^2)(\frac{dy}{dx}) = 3x^2 + 1$ . To isolate $(\frac{dy}{dx})$ , divide both sides by $(-4y - 6y^2)$ .
Isolate $(\frac{dy}{dx}):</b> Now we have \$(-4y - 6y^2)(\frac{dy}{dx}) = 3x^2 + 1$ . To isolate $(\frac{dy}{dx})$ , divide both sides by $(-4y - 6y^2)$ .After dividing, we get $(\frac{dy}{dx}) = \frac{3x^2 + 1}{-4y - 6y^2}$ .

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