Implicit differentiation

Examples

Find $\frac{d y}{d x}$ of $x^{2} + x y + y^{2} = 1$ .

\begin{aligned} x^{2} + x y + y^{2} & = 1 \\ \frac{d}{d x} (x^{2} + x y + y^{2}) & = \frac{d}{d x} 1 \\ 2 x + y + x y^{'} + 2 y y^{'} & = 0 \\ y^{'} (x + 2 y) & = - (2 x + y) \\ y^{'} & = - \frac{2 x + y}{x + 2 y} \end{aligned}

Find $\frac{d x}{d y}$ of $x^{2} + x y + y^{2} = 1$ .

\begin{aligned} x^{2} + x y + y^{2} & = 1 \\ \frac{d}{d x} (x^{2} + x y + y^{2}) & = \frac{d}{d x} 1 \\ 2 x x^{'} + x^{'} y + x + 2 y & = 0 \\ x^{'} (2 x + y) & = - (x + 2 y) \\ x^{'} & = - \frac{x + 2 y}{2 x + y} \end{aligned}

Find the slope of the tangent line to the curve $x y \ln (2 - y x) = 0$ at $(1, 1)$ .

\begin{aligned} x y \ln (2 - x y) & = 0 \\ \frac{d}{d x} (x y \ln (2 - x y)) & = \frac{d}{d x} 0 \\ y \ln (2 - x y) + x y^{'} \ln (2 - x y) + x y \cdot \frac{(2 - x y)^{'}}{2 - x y} & = 0 \\ y \ln (2 - x y) + x y^{'} \ln (2 - x y) - x y \cdot \frac{y + x y^{'}}{2 - x y} & = 0 \end{aligned}

Plug $(1, 1)$ :

\begin{aligned} 1 \ln (2 - 1 \cdot 1) + 1 y^{'} (1, 1) \ln (2 - 1 \cdot 1) - 1 \cdot 1 \frac{1 + 1 y^{'} (1, 1)}{2 - 1 \cdot 1} & = 0 \\ - (1 + y^{'} (1, 1)) & = 0 \\ y^{'} (1, 1) & = - 1 \end{aligned}

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