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[tex]y+2=(x^2/2)-2sin(y)[/tex]

Sagot :

the assumption being that "y" is encapsulating a function in "x" terms.

[tex]y+2=\cfrac{x^2}{2}-2sin(y)\implies \stackrel{\textit{taking the derivative to both sides}}{\cfrac{dy}{dx}+0=\cfrac{1}{2}(2x)\underset{chain~rule}{-2cos(y)\cdot \cfrac{dy}{dx}}} \\\\\\ \cfrac{dy}{dx}=x-2cos(y)\cfrac{dy}{dx}\implies \cfrac{dy}{dx}+2cos(y)\cfrac{dy}{dx}=x \\\\\\ \cfrac{dy}{dx}\left[ 1+2cos(y) \right]=x\implies \cfrac{dy}{dx}=\cfrac{x}{1+2cos(y)}[/tex]

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