Calculating derivatives Find dy&gt;dx for the following functions. y=12+sin⁡x

Step 1Calculating derivatives Find dydx for the following functions.y=12+sin⁡xStep 2We have, y=12+sin⁡x…(1)Differentiate equation(1) w.r.t 'x' , we getApply quotient rule :ddx[f(x)g(x)]=g(x)f′(x)−f(x)g′(x)[g(x)]2⇒dydx=(2sin⁡x)ddx(1)−1.ddx(2+sin⁡x)(2+sin⁡x)2⇒dydx=(2+sin⁡x)×0−cos⁡x(2+sin⁡x)2,(ddx(sin⁡x)=cos⁡x)⇒dydx=−cos⁡x(2+sin⁡x)2Hence, dydx=−cos⁡x(2+sin⁡x)2

"Calculating derivatives Find dy>dx for the following functions...." solved and explained

Zoe Oneal

Answered question

2021-05-11

Calculating derivatives Find dy>dx for the following functions.

y = \frac{1}{2 + \sin x}

Answer & Explanation

Roosevelt Houghton

Skilled2021-05-12Added 106 answers

Step 1
Calculating derivatives Find $\frac{d y}{d x}$ for the following functions.
$y = \frac{1}{2 + \sin x}$
Step 2
We have, $y = \frac{1}{2 + \sin x} \dots$ (1)
Differentiate equation(1) w.r.t 'x' , we get
Apply quotient rule : $\frac{d}{d x} [\frac{f (x)}{g (x)}] = \frac{g (x) f^{'} (x) - f (x) g^{'} (x)}{{[g (x)]}^{2}}$
$\Rightarrow \frac{d y}{d x} = \frac{(2 \sin x) \frac{d}{d x} (1) - 1 . \frac{d}{d x} (2 + \sin x)}{{(2 + \sin x)}^{2}}$
$\Rightarrow \frac{d y}{d x} = \frac{(2 + \sin x) \times 0 - \cos x}{{(2 + \sin x)}^{2}}, (\frac{d}{d x} (\sin x) = \cos x)$
$\Rightarrow \frac{d y}{d x} = \frac{- \cos x}{{(2 + \sin x)}^{2}}$
Hence, $\frac{d y}{d x} = \frac{- \cos x}{{(2 + \sin x)}^{2}}$

Jeffrey Jordon

Expert2021-11-24Added 2605 answers

Answer is given below (on video)

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