If -2x-5y^(2)+4+5x^(3)=0 then find (dy)/(dx) in terms of x and y. Answer: (dy)/(dx)=

Given Equation: We are given the equation -2x - 5y^2 + 4 + 5x^3 = 0 and we need to find the derivative of y with respect to x, which is denoted as (dy/dx). To do this, we will use implicit differentiation, which involves taking the derivative of both sides of the equation with respect to x, while treating y as a function of x.Implicit Differentiation: First, we differentiate each term of the equation with respect to x. The derivative of -2x with respect to x is -2. The derivative of -5y^2 with respect to x is -10y(dy/dx), using the chain rule because y is a function of x. The derivative of 4 with respect to x is 0, since it is a constant. The derivative of 5x^3 with respect to x is 15x^2.Differentiating Terms: Now we write down the differentiated equation: -2 - 10y(dy/dx) + 0 + 15x^2 = 0.Write Down Differentiated Equation: Next, we solve for (dy/dx). To do this, we isolate the term containing (dy/dx) on one side of the equation. We get -10y(dy/dx) = 2 - 15x^2.Solve for (dy/dx): Now, we divide both sides of the equation by -10y to solve for (dy/dx). This gives us (dy/dx) = (2 - 15x^2) / (-10y).Divide to Solve for (dy/dx): We have found the derivative of y with respect to x in terms of x and y, which is (dy/dx) = (2 - 15x^2) / (-10y).

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If
-2x-5y^(2)+4+5x^(3)=0 then find
(dy)/(dx) in terms of
x and
y.
Answer:
(dy)/(dx)=

If $-2 x-5 y^{2}+4+5 x^{3}=0$ then find $\frac{d y}{d x}$ in terms of $x$ and $y$ . $\newline$ Answer: $\frac{d y}{d x}=$

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Q. If

-2 x-5 y^{2}+4+5 x^{3}=0

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 $-2x - 5y^2 + 4 + 5x^3 = 0$ and we need to find the derivative of $y$ with respect to $x$ , which is denoted as $\frac{dy}{dx}$ . To do this, we will use implicit differentiation, which involves taking the derivative of both sides of the equation with respect to $x$ , while treating $y$ as a function of $x$ .
Implicit Differentiation: First, we differentiate each term of the equation with respect to $x$ . The derivative of $-2x$ with respect to $x$ is $-2$ . The derivative of $-5y^2$ with respect to $x$ is $-10y(\frac{dy}{dx})$ , using the chain rule because $y$ is a function of $x$ . The derivative of $4$ with respect to $x$ is $-2x$ $1$ , since it is a constant. The derivative of $-2x$ $2$ with respect to $x$ is $-2x$ $4$ .
Differentiating Terms: Now we write down the differentiated equation: $-2 - 10y\frac{dy}{dx} + 0 + 15x^2 = 0$ .
Write Down Differentiated Equation: Next, we solve for $\frac{dy}{dx}$ . To do this, we isolate the term containing $\frac{dy}{dx}$ on one side of the equation. We get $-10y\frac{dy}{dx} = 2 - 15x^2$ .
Solve for $(\frac{dy}{dx}):</b> Now, we divide both sides of the equation by \$-10y$ to solve for $(\frac{dy}{dx})$ . This gives us $(\frac{dy}{dx}) = \frac{2 - 15x^2}{-10y}$ .
Divide to Solve for $(\frac{dy}{dx}):$ We have found the derivative of $y$ with respect to $x$ in terms of $x$ and $y$ , which is $(\frac{dy}{dx}) = \frac{2 - 15x^2}{-10y}$ .

If $-2 x-5 y^{2}+4+5 x^{3}=0$ then find $\frac{d y}{d x}$ in terms of $x$ and $y$ . $\newline$ Answer: $\frac{d y}{d x}=$

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If $-2 x-5 y^{2}+4+5 x^{3}=0$ then find $\frac{d y}{d x}$ in terms of $x$ and $y$ . $\newline$ Answer: $\frac{d y}{d x}=$