find $\\frac{dy}{dx}$ by implicit differentiation.\n$y = 4x^{2}+9yx$\n$\\frac{dy}{dx}=\\frac{8x - 9y}{1…

find $\\frac{dy}{dx}$ by implicit differentiation.\n$y = 4x^{2}+9yx$\n$\\frac{dy}{dx}=\\frac{8x - 9y}{1 - 9x}$\nthats incorrect.\nto use implicit differentiation, first differentiate both sides of the equation with\nrespect to $x$, treating $y$ as a differentiable function of $x$. then, collect the terms with\n$\\frac{dy}{dx}$ on one side of the equation. finally, solve for $\\frac{dy}{dx}$.

find $\\frac{dy}{dx}$ by implicit differentiation.\n$y = 4x^{2}+9yx$\n$\\frac{dy}{dx}=\\frac{8x - 9y}{1 - 9x}$\nthats incorrect.\nto use implicit differentiation, first differentiate both sides of the equation with\nrespect to $x$, treating $y$ as a differentiable function of $x$. then, collect the terms with\n$\\frac{dy}{dx}$ on one side of the equation. finally, solve for $\\frac{dy}{dx}$.

Answer

Explanation:

Step1: Differentiate both sides

Differentiate (y = 4x^{2}+9yx) with respect to (x). Using the sum rule ((u + v)^\prime=u^\prime + v^\prime), where (u = 4x^{2}) and (v=9yx). The derivative of (4x^{2}) with respect to (x) is (8x) (using the power rule ((x^{n})^\prime=nx^{n - 1})). For the term (9yx), use the product rule ((uv)^\prime=u^\prime v+uv^\prime), where (u = 9y) and (v=x). So ((9yx)^\prime=9\frac{dy}{dx}\cdot x+9y\cdot1 = 9x\frac{dy}{dx}+9y). The derivative of (y) with respect to (x) is (\frac{dy}{dx}). So we have (\frac{dy}{dx}=8x + 9x\frac{dy}{dx}+9y).

Step2: Collect (\frac{dy}{dx}) terms

Subtract (9x\frac{dy}{dx}) from both sides: (\frac{dy}{dx}-9x\frac{dy}{dx}=8x + 9y). Factor out (\frac{dy}{dx}) on the left - hand side: (\frac{dy}{dx}(1 - 9x)=8x + 9y).

Step3: Solve for (\frac{dy}{dx})

Divide both sides by ((1 - 9x)) (assuming (x\neq\frac{1}{9})): (\frac{dy}{dx}=\frac{8x + 9y}{1 - 9x}).

Answer:

(\frac{dy}{dx}=\frac{8x + 9y}{1 - 9x})