Compute the following derivatives

$\frac{d}{dx}\left[\frac{\mathrm{sin}x}{1+\mathrm{cos}x}\right]$

Tobias Ali
2021-05-21
Answered

Compute the following derivatives

$\frac{d}{dx}\left[\frac{\mathrm{sin}x}{1+\mathrm{cos}x}\right]$

You can still ask an expert for help

Isma Jimenez

Answered 2021-05-22
Author has **84** answers

Step 1

To find the derivatives

Step 2

$\left(\frac{d}{dx}\right)\left(\frac{\mathrm{sin}x}{1+\mathrm{cos}x}\right)$

Differentiating with respect to x then

$y}^{\prime}\frac{(1+\mathrm{cos}x)\times \mathrm{cos}x-\mathrm{sin}x(-\mathrm{sin}x)}{{(1+\mathrm{cos}x)}^{2}$

Since, used quotient rule

$\left(\frac{u}{v}\right)}^{\prime}=\frac{v{u}^{\prime}-u{v}^{\prime}}{{v}^{2}$

$y}^{\prime}=\frac{\mathrm{cos}x+{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x}{{(1+\mathrm{cos}x)}^{2}$

Since,${\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1$ so,

$y}^{\prime}=\frac{1+\mathrm{cos}x}{{(1+\mathrm{cos}x)}^{2}$

$y}^{\prime}={(1+\mathrm{cos}x)}^{1-2$

Since,$\frac{{x}^{a}}{{x}^{b}}={x}^{a-b}$

$y}^{\prime}={(1+\mathrm{cos}x)}^{-1$

$y}^{\prime}=\frac{1}{1+\mathrm{cos}x$

To find the derivatives

Step 2

Differentiating with respect to x then

Since, used quotient rule

Since,

Since,

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