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• Notes

For the following functions find $\frac{d y}{d x}$

$x^{3}+y^{3}=1$
$\mathbf{e}^{x / y}=x y$
$\frac{1}{x}+\frac{1}{y}=2$

$x^{3}+y^{3}=1 \rightarrow 3 x^{2}+3 y^{2} \frac{d y}{d x}=0 \rightarrow x^{2}+y^{2} \frac{d y}{d x}=0$

$\frac{d y}{d x}=\frac{-x^{2}}{y^{2}}$

$e^{x / y}=x-y \longrightarrow e^{x / y}\left(\frac{\left(1)(y)-\frac{d y}{d x}(x)\right.}{y^{2}}\right)=1-\frac{d y}{d x}$

$e^{x / y}\left(y-x \frac{d y}{d x}\right)=y^{2}-y^{2} \frac{d y}{d x}$

$\rightarrow y e^{x / y}-x e^{x / y} \frac{d y}{d x}=y^{2}-y^{2} \frac{d y}{d x}$

$-x e^{x / y} \frac{d y}{d x}+y^{2} \frac{d y}{d x}=y^{2}-y e^{x / y}$

$\frac{d y}{d x}=\frac{y^{2}-y e^{x / y}}{-x e^{x / y}+y^{2}}$

$\frac{1}{x}+\frac{1}{y}=2 \longrightarrow \frac{-1}{x^{2}}+\frac{-\frac{d y}{d x}}{y^{2}}=0$

$\frac{\frac{-d y}{d x}}{y^{2}}=1 / x^{2} \rightarrow$

$\frac{-d y}{d x}=\frac{y^{2}}{x^{2}} \rightarrow \frac{d x}{d x}=\frac{-y^{2}}{x^{2}}$

find $\frac{d y}{d x}$ by implicit differentiation:

$y \cos x=x^{2}+y^{2}$

$y \cos x=x^{2}+y^{2} \rightarrow \frac{d y}{d x}(\cos x)+(-\sin x)(y)=2 x+2 y \frac{d y}{d x}$

$\frac{d y}{d x} \cos x-2 y \frac{d y}{d x}=2 x+y \sin x$

$\frac{d y}{d x}=\frac{2 x+y \sin x}{\cos x-2 y}$