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

Differentiate with chain rule: First, we need to differentiate both sides of the equation with respect to x. The left side of the equation involves y, which is a function of x, so we will use the chain rule to differentiate it. The right side of the equation only involves x, so we will differentiate it normally.Differentiate left side: Differentiate the left side of the equation with respect to x: d/dx (5y^(3) + y). Using the chain rule, this becomes 15y^(2)(dy/dx) + (dy/dx).Differentiate right side: Differentiate the right side of the equation with respect to x: d/dx (-y^(2) - 2x^(2) + x^(3)). This becomes -2(dy/dx)y - 4x + 3x^(2).Combine differentiated terms: Now we have the following equation after differentiation: 15y^(2)(dy/dx) + (dy/dx) = -2(dy/dx)y - 4x + 3x^(2).Factor out common term: Combine the terms involving (dy/dx) on one side of the equation: 15y^(2)(dy/dx) + (dy/dx) + 2(dy/dx)y = 3x^(2) - 4x.Solve for dy/dx: Factor out (dy/dx) from the left side of the equation: (dy/dx)(15y^(2) + 1 + 2y) = 3x^(2) - 4x.Solve for (dy/dx): (dy/dx) = (3x^(2) - 4x) / (15y^(2) + 1 + 2y).

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

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

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Math Problems

Algebra 2

Evaluate exponential functions

Full solution

Q. If

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

then find

\frac{d y}{d x}

in terms of

x

and

y

\newline

Answer:

\frac{d y}{d x}=

Differentiate with chain rule: First, we need to differentiate both sides of the equation with respect to $x$ . The left side of the equation involves $y$ , which is a function of $x$ , so we will use the chain rule to differentiate it. The right side of the equation only involves $x$ , so we will differentiate it normally.
Differentiate left side: Differentiate the left side of the equation with respect to $x$ : $\frac{d}{dx} (5y^{3} + y)$ . Using the chain rule, this becomes $15y^{2}(\frac{dy}{dx}) + (\frac{dy}{dx})$ .
Differentiate right side: Differentiate the right side of the equation with respect to $x$ : $\frac{d}{dx} (-y^{2} - 2x^{2} + x^{3})$ . This becomes $-2\frac{dy}{dx}y - 4x + 3x^{2}$ .
Combine differentiated terms: Now we have the following equation after differentiation: $15y^{2}\left(\frac{dy}{dx}\right) + \left(\frac{dy}{dx}\right) = -2\left(\frac{dy}{dx}\right)y - 4x + 3x^{2}.$
Factor out common term: Combine the terms involving $\frac{dy}{dx}$ on one side of the equation: $15y^{2}\left(\frac{dy}{dx}\right) + \left(\frac{dy}{dx}\right) + 2\left(\frac{dy}{dx}\right)y = 3x^{2} - 4x$ .
Solve for $\frac{dy}{dx}$ : Factor out $\frac{dy}{dx}$ from the left side of the equation: $\left(\frac{dy}{dx}\right)(15y^{2} + 1 + 2y) = 3x^{2} - 4x$ .
Solve for $\frac{dy}{dx}$ : Factor out $\frac{dy}{dx}$ from the left side of the equation: $\left(\frac{dy}{dx}\right)(15y^{2} + 1 + 2y) = 3x^{2} - 4x$ . Solve for $\frac{dy}{dx}$ : $\frac{dy}{dx} = \frac{3x^{2} - 4x}{15y^{2} + 1 + 2y}$ .