Question

if 2x + y = - cosx then dy /dx is equals to​1.​

Answer 1

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:

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

$\therefore$

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

Answer 2

Answer:

Answer:

Answer:

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

dx

dy

=sin x−2

Given:

\sf 2x + y = -cos \: x2x+y=−cosx

To Find:

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

dx

dy

i.e.y

′

(x)

Step-by-step explanation:

\begin{gathered}\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\end{gathered}

Findthederivativeofthefollowing

viaimplicitdifferentiation:

⟹

dx

d

(2x+y)=

dx

d

(−cosx)

Differentiatethesumtermbyterm

andfactoroutconstants:

⟹2

dx

d

(x)+

dx

d

(y)=−

dx

d

(cosx)

Thederivativeofxis1:

⟹2×1+

dx

dy

=−

dx

d

(cosx)

dx

d

(cosx)=−sinx:

⟹2+

dx

dy

=−(−sinx)

⟹2+

dx

dy

=sinx

⟹

dx

dy

=sinx−2

\therefore∴

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

dx

dy

i.e.y

′

(x)=sinx−2

Step-by-step explanation:

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