Question

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Answer 1

Answer:

Answer:

Answer:

$\boxed{\mathfrak{\frac{dy}{dx} = sin \ x - 2}}$

Given:

$\sf 2x + y = -cos \: x$

To Find:

$\sf \frac{dy}{dx} \: i.e. \: y'(x)$

Step-by-step explanation:

$\sf Find \: the \: derivative \: of \: the \: following \\ \sf via \: implicit \: differentiation: \\ \sf \implies \frac{d}{dx} (2x + y) = \frac{d}{dx} ( - cos \: x) \\ \\ \sf Differentiate \: the \: sum \: term \: by \: term \\ \sf and \: factor \: out \: constants: \\ \sf \implies 2 \frac{d}{dx} (x) + \frac{d}{dx} (y) = - \frac{d}{dx} ( cos \: x) \\ \\ \sf The \: derivative \: of \: x \: is \: 1: \\ \sf \implies 2 \times 1 + \frac{dy}{dx} = - \frac{d}{dx} ( cos \: x) \\ \\ \sf \frac{d}{dx} ( cos \: x) = - sin \: x : \\ \sf \implies 2 + \frac{dy}{dx} = - ( - sin \: x) \\ \\ \sf \implies 2 + \frac{dy}{dx} = sin \: x \\ \\ \sf \implies \frac{dy}{dx} = sin \: x - 2$

$\therefore$

$\sf \frac{dy}{dx} \: i.e. \: y'(x) = sin \: x - 2$

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Answer 2

Answer:

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