Full solution

Q. Using implicit differentiation, find

\frac{d y}{d x}

\newline

-7 y^{2}-2 x^{3}+3 y+6 x-x y=-7

Given Equation: Given the equation: $-7y^{2} - 2x^{3} + 3y + 6x - xy = -7$ , we need to find the derivative of $y$ with respect to $x$ , denoted as $\frac{dy}{dx}$ , using implicit differentiation. $\newline$ To differentiate each term with respect to $x$ , we apply the rules of differentiation, keeping in mind that $y$ is a function of $x$ .
Differentiate $y^2$ : Differentiate $-7y^{2}$ with respect to $x$ . Since $y$ is a function of $x$ , we use the chain rule: the derivative of $y^{2}$ is $2y\frac{dy}{dx}$ , and we multiply this by the constant $-7$ . $\newline$ The derivative of $-7y^{2}$ is $-14y\frac{dy}{dx}$ .
Differentiate $x^3$ : Differentiate $-2x^{3}$ with respect to $x$ . Since $x$ is the variable we are differentiating with respect to, the derivative is simply the power rule: $-2 \times 3x^{2}$ . $\newline$ The derivative of $-2x^{3}$ is $-6x^{2}$ .
Differentiate $3y$ : Differentiate $3y$ with respect to $x$ . Again, since $y$ is a function of $x$ , the derivative is $3\frac{dy}{dx}$ .
Differentiate $6x$ : Differentiate $6x$ with respect to $x$ . This is a simple differentiation, and the derivative is $6$ .
Differentiate $-xy$ : Differentiate $-xy$ with respect to $x$ . This requires the product rule since both $x$ and $y$ are functions of $x$ . The product rule states that $\frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ , where $u = x$ and $v = y$ . $\newline$ The derivative of $-xy$ is $-xy$ $0$ .
Differentiate $-7$ : Differentiate $-7$ with respect to $x$ . Since $-7$ is a constant, its derivative is $0$ .
Combine Differentiated Terms: Combine all the differentiated terms to form the derivative of the entire left side of the equation with respect to $x$ . $-14y\frac{dy}{dx} - 6x^{2} + 3\frac{dy}{dx} + 6 - y - x\frac{dy}{dx} = 0$
Collect Terms: Now, we collect all terms involving $\frac{dy}{dx}$ on one side and the rest on the other side to solve for $\frac{dy}{dx}$ . $\left(-14y - x\right)\frac{dy}{dx} + 3\frac{dy}{dx} = 6x^{2} + y - 6$
Factor Out $(dy)/(dx)$ : Factor out $(dy)/(dx)$ from the terms on the left side of the equation. $\newline$ (dy)/(dx)(\(-14y - x + $3$ ) = $6$ x^{ $2$ } + y - $6$
Solve for $(\frac{dy}{dx}):$ Solve for $(\frac{dy}{dx})$ by dividing both sides of the equation by $(-14y - x + 3)$ . $(\frac{dy}{dx}) = \frac{6x^{2} + y - 6}{-14y - x + 3}$