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

Given Equation: We are given the equation 0 = x^3 - y + 3 - 5y^3. 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 Left Side: Differentiate the left side of the equation with respect to x. The derivative of 0 with respect to x is 0.Differentiate Right Side: Differentiate the right side of the equation with respect to x. The derivative of x^3 with respect to x is 3x^2. The derivative of -y with respect to x is -dy/dx (since y is a function of x). The derivative of the constant 3 with respect to x is 0. The derivative of -5y^3 with respect to x is -15y^2 * dy/dx (using the chain rule).Combine Derivatives: Combine the derivatives to form the differentiated equation: 0 = 3x^2 - dy/dx - 15y^2 * dy/dx.Rearrange Equation: Rearrange the equation to solve for dy/dx: dy/dx + 15y^2 * dy/dx = 3x^2.Factor out dy/dx: Factor out dy/dx on the left side of the equation: dy/dx * (1 + 15y^2) = 3x^2.Isolate dy/dx: Divide both sides of the equation by (1 + 15y^2) to isolate dy/dx: dy/dx = 3x^2 / (1 + 15y^2).

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

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

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

then find

\frac{d y}{d x}

in terms of

x

and

y

\newline

Answer:

\frac{d y}{d x}=

Given Equation: We are given the equation $0 = x^3 - y + 3 - 5y^3$ . 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 Left Side: Differentiate the left side of the equation with respect to $x$ . The derivative of $0$ with respect to $x$ is $0$ .
Differentiate Right Side: Differentiate the right side of the equation with respect to $x$ . The derivative of $x^3$ with respect to $x$ is $3x^2$ . The derivative of $-y$ with respect to $x$ is $-\frac{dy}{dx}$ (since $y$ is a function of $x$ ). The derivative of the constant $3$ with respect to $x$ is $x^3$ $1$ . The derivative of $x^3$ $2$ with respect to $x$ is $x^3$ $4$ (using the chain rule).
Combine Derivatives: Combine the derivatives to form the differentiated equation: $0 = 3x^2 - \frac{dy}{dx} - 15y^2 \cdot \frac{dy}{dx}$ .
Rearrange Equation: Rearrange the equation to solve for $dy/dx$ : $\frac{dy}{dx} + 15y^2 \frac{dy}{dx} = 3x^2$ .
Factor out dy/dx: Factor out $\frac{dy}{dx}$ on the left side of the equation: $\frac{dy}{dx} \times (1 + 15y^2) = 3x^2$ .
Isolate $\frac{dy}{dx}$ : Divide both sides of the equation by $(1 + 15y^2)$ to isolate $\frac{dy}{dx}$ : $\frac{dy}{dx} = \frac{3x^2}{(1 + 15y^2)}$ .