((y)/(x)+xy=-2)(d)/(dx)

Given Equation Differentiation: We are given the equation (y/x) + xy = -2 and we need to find its derivative with respect to x. We will use the rules of differentiation to differentiate each term separately.Quotient Rule Application: Differentiate the first term (y/x) with respect to x. Since y is a function of x, we will use the quotient rule which is d(u/v)/dx = (v(du/dx) - u(dv/dx))/v^2. Here, u = y and v = x, so du/dx = dy/dx and dv/dx = 1.Product Rule Application: Applying the quotient rule to y/x, we get: d(y/x)/dx = (x(dy/dx) - y(1))/x^2 = (x(dy/dx) - y)/x^2Combining Derivatives: Differentiate the second term xy with respect to x. Since this is a product of two functions, we will use the product rule which is d(uv)/dx = u(dv/dx) + v(du/dx). Here, u = x and v = y, so du/dx = 1 and dv/dx = dy/dx.Simplifying Equation: Applying the product rule to xy, we get: d(xy)/dx = x(dy/dx) + y(1) = x(dy/dx) + yCorrection of Mistake: Now, we combine the derivatives of both terms and set the derivative of the entire equation equal to the derivative of -2, which is 0 since -2 is a constant. (d(y/x)/dx) + d(xy)/dx = d(-2)/dx ((x(dy/dx) - y)/x^2) + (x(dy/dx) + y) = 0We simplify the equation by combining like terms. However, we need to be careful with the terms involving dy/dx, as they cannot be simply added or subtracted without considering their coefficients. ((x(dy/dx) - y)/x^2) + (x(dy/dx) + y) = 0 (x(dy/dx) - y + x^2(dy/dx) + x^2*y)/x^2 = 0We made a mistake in the previous step by incorrectly combining the terms. We should have found a common denominator before combining them. Let's correct this. (x(dy/dx) - y)/x^2 + x(dy/dx) + y = 0 (x(dy/dx) - y + x^3(dy/dx) + x^2*y)/x^2 = 0

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Algebra 1

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\left(\frac{y}{x}+xy=-2\right)\frac{d}{dx}

Given Equation Differentiation: We are given the equation $(\frac{y}{x}) + xy = -2$ and we need to find its derivative with respect to $x$ . We will use the rules of differentiation to differentiate each term separately.
Quotient Rule Application: Differentiate the first term $\frac{y}{x}$ with respect to $x$ . Since $y$ is a function of $x$ , we will use the quotient rule which is $\frac{d(\frac{u}{v})}{dx} = \frac{v(\frac{du}{dx}) - u(\frac{dv}{dx})}{v^2}$ . Here, $u = y$ and $v = x$ , so $\frac{du}{dx} = \frac{dy}{dx}$ and $\frac{dv}{dx} = 1$ .
Product Rule Application: Applying the quotient rule to $\frac{y}{x}$ , we get: $\newline$ $\frac{d(\frac{y}{x})}{dx} = \frac{x(\frac{dy}{dx}) - y(1)}{x^2}$ $\newline$ $= \frac{x(\frac{dy}{dx}) - y}{x^2}$
Combining Derivatives: Differentiate the second term $xy$ with respect to $x$ . Since this is a product of two functions, we will use the product rule which is $\frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ . Here, $u = x$ and $v = y$ , so $\frac{du}{dx} = 1$ and $\frac{dv}{dx} = \frac{dy}{dx}$ .
Simplifying Equation: Applying the product rule to $xy$ , we get: $\frac{d(xy)}{dx} = x\left(\frac{dy}{dx}\right) + y(1)$ $= x\left(\frac{dy}{dx}\right) + y$
Correction of Mistake: Now, we combine the derivatives of both terms and set the derivative of the entire equation equal to the derivative of $-2$ , which is $0$ since $-2$ is a constant. $\newline$ $\frac{d(\frac{y}{x})}{dx} + \frac{d(xy)}{dx} = \frac{d(-2)}{dx}$ $\newline$ $\left(\frac{x(\frac{dy}{dx}) - y}{x^2}\right) + (x(\frac{dy}{dx}) + y) = 0$
Correction of Mistake: Now, we combine the derivatives of both terms and set the derivative of the entire equation equal to the derivative of $-2$ , which is $0$ since $-2$ is a constant. $\newline$ $\frac{d(\frac{y}{x})}{dx} + \frac{d(xy)}{dx} = \frac{d(-2)}{dx}$ $\newline$ $\frac{(x\frac{dy}{dx} - y)}{x^2} + (x\frac{dy}{dx} + y) = 0$ We simplify the equation by combining like terms. However, we need to be careful with the terms involving $\frac{dy}{dx}$ , as they cannot be simply added or subtracted without considering their coefficients. $\newline$ $\frac{(x\frac{dy}{dx} - y)}{x^2} + (x\frac{dy}{dx} + y) = 0$ $\newline$ $\frac{(x\frac{dy}{dx} - y + x^2\frac{dy}{dx} + x^2*y)}{x^2} = 0$
Correction of Mistake: Now, we combine the derivatives of both terms and set the derivative of the entire equation equal to the derivative of $-2$ , which is $0$ since $-2$ is a constant. $\newline$ $\frac{d(\frac{y}{x})}{dx} + \frac{d(xy)}{dx} = \frac{d(-2)}{dx}$ $\newline$ $\left(\frac{x(\frac{dy}{dx}) - y}{x^2}\right) + (x(\frac{dy}{dx}) + y) = 0$ We simplify the equation by combining like terms. However, we need to be careful with the terms involving $\frac{dy}{dx}$ , as they cannot be simply added or subtracted without considering their coefficients. $\newline$ $\left(\frac{x(\frac{dy}{dx}) - y}{x^2}\right) + (x(\frac{dy}{dx}) + y) = 0$ $\newline$ $\frac{x(\frac{dy}{dx}) - y + x^2(\frac{dy}{dx}) + x^2\cdot y}{x^2} = 0$ We made a mistake in the previous step by incorrectly combining the terms. We should have found a common denominator before combining them. Let's correct this. $\newline$ $\frac{x(\frac{dy}{dx}) - y}{x^2} + x(\frac{dy}{dx}) + y = 0$ $\newline$ $\frac{x(\frac{dy}{dx}) - y + x^3(\frac{dy}{dx}) + x^2\cdot y}{x^2} = 0$