If -y+5x^(2)+y^(3)-5=0 then find (dy)/(dx) in terms of x and y. Answer: (dy)/(dx)=

Differentiate -y: We are given the equation -y + 5x^2 + y^3 - 5 = 0. To find (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 5x^2: Differentiate -y with respect to x. Since y is a function of x, we use the chain rule and get -dy/dx.Differentiate y^3: Differentiate 5x^2 with respect to x. The derivative of 5x^2 with respect to x is 10x.Differentiate -5: Differentiate y^3 with respect to x. Again, using the chain rule, the derivative is 3y^2 * dy/dx.Combine differentiated terms: Differentiate -5 with respect to x. The derivative of a constant is 0.Solve for dy/dx: Combine all the differentiated terms to rewrite the equation: -dy/dx + 10x + 3y^2 * dy/dx - 0 = 0.Factor out dy/dx: Now, we need to solve for dy/dx. Group all the terms containing dy/dx on one side and the rest on the other side: dy/dx (-1 + 3y^2) = -10x.Isolate dy/dx: Factor out dy/dx from the left side: dy/dx * (-1 + 3y^2) = -10x.Simplify dy/dx: Divide both sides by (-1 + 3y^2) to isolate dy/dx: dy/dx = -10x / (-1 + 3y^2).Simplify the expression for dy/dx: dy/dx = 10x / (1 - 3y^2).

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If
-y+5x^(2)+y^(3)-5=0 then find
(dy)/(dx) in terms of
x and
y.
Answer:
(dy)/(dx)=

If $-y+5 x^{2}+y^{3}-5=0$ 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 2

Evaluate exponential functions

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Q. If

-y+5 x^{2}+y^{3}-5=0

then find

\frac{d y}{d x}

in terms of

x

and

y

\newline

Answer:

\frac{d y}{d x}=

Differentiate $-y$ : We are given the equation $-y + 5x^2 + y^3 - 5 = 0$ . 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 $5x^2$ : Differentiate $-y$ with respect to $x$ . Since $y$ is a function of $x$ , we use the chain rule and get $-\frac{dy}{dx}$ .
Differentiate $y^3$ : Differentiate $5x^2$ with respect to $x$ . The derivative of $5x^2$ with respect to $x$ is $10x$ .
Differentiate $-5$ : Differentiate $y^3$ with respect to $x$ . Again, using the chain rule, the derivative is $3y^2 \cdot \frac{dy}{dx}$ .
Combine differentiated terms: Differentiate $-5$ with respect to $x$ . The derivative of a constant is $0$ .
Solve for $\frac{dy}{dx}$ : Combine all the differentiated terms to rewrite the equation: $-\frac{dy}{dx} + 10x + 3y^2 \cdot \frac{dy}{dx} - 0 = 0$ .
Factor out $\frac{dy}{dx}$ : Now, we need to solve for $\frac{dy}{dx}$ . Group all the terms containing $\frac{dy}{dx}$ on one side and the rest on the other side: $\frac{dy}{dx} (-1 + 3y^2) = -10x$ .
Isolate $\frac{dy}{dx}$ : Factor out $\frac{dy}{dx}$ from the left side: $\frac{dy}{dx} \times (-1 + 3y^2) = -10x$ .
Simplify $\frac{dy}{dx}$ : Divide both sides by $(-1 + 3y^2)$ to isolate $\frac{dy}{dx}$ : $\frac{dy}{dx} = \frac{-10x}{(-1 + 3y^2)}$ .
Simplify $\frac{dy}{dx}$ : Divide both sides by $(-1 + 3y^2)$ to isolate $\frac{dy}{dx}$ : $\frac{dy}{dx} = \frac{-10x}{(-1 + 3y^2)}$ .Simplify the expression for $\frac{dy}{dx}$ : $\frac{dy}{dx} = \frac{10x}{(1 - 3y^2)}$ .