Find dy/dx in terms of \( x \& y \) : \( \begin{array}{l}\text { 1. } x y-x+2 y=1 \\ \text { 2. } x^{3}+y^{3}=1 \\ \text { 3. } x^{2}+x y=y^{3} \\ \text { 4. } x^{3} y+x y^{5}=2 \\ \text { 5. } x^{2} y^{3}=2 x-y \\ \text { 6. } x^{2}+4(y-1)^{2}=4 \\ \text { 7. } \frac{x-y}{x+y}=\frac{x^{2}}{y}+1\end{array} \)
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Sure! Let’s explore how to differentiate these implicit functions step by step. For the first equation, \( xy - x + 2y = 1 \), you can implicitly differentiate both sides with respect to \( x \). Applying the product rule and moving terms involving \( dy/dx \) will give you the result: \[ y + x \frac{dy}{dx} - 1 + 2 \frac{dy}{dx} = 0 \] Rearranging gives you \( \frac{dy}{dx} = \frac{1 - y}{x + 2} \). Now let’s jump to the fifth equation, \(x^2y^3 = 2x - y\). Again, use implicit differentiation: \[ 2xy^3 + 3x^2y^2\frac{dy}{dx} = 2 - \frac{dy}{dx} \] After rearranging and solving for \( \frac{dy}{dx} \), you would find \[ \frac{dy}{dx} = \frac{2 - 2xy^3}{3x^2y^2 + 1} \] And there we have it! You're ready to tackle these implicit derivatives like a pro.