# Compute the following derivatives \frac{d}{dx}[\frac{\sin x}{1+\cos x}]

Tobias Ali 2021-05-21 Answered
Compute the following derivatives
$\frac{d}{dx}\left[\frac{\mathrm{sin}x}{1+\mathrm{cos}x}\right]$
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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
${y}^{\prime }\frac{\left(1+\mathrm{cos}x\right)×\mathrm{cos}x-\mathrm{sin}x\left(-\mathrm{sin}x\right)}{{\left(1+\mathrm{cos}x\right)}^{2}}$
Since, used quotient rule
${\left(\frac{u}{v}\right)}^{\prime }=\frac{v{u}^{\prime }-u{v}^{\prime }}{{v}^{2}}$
${y}^{\prime }=\frac{\mathrm{cos}x+{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x}{{\left(1+\mathrm{cos}x\right)}^{2}}$
Since, ${\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1$ so,
${y}^{\prime }=\frac{1+\mathrm{cos}x}{{\left(1+\mathrm{cos}x\right)}^{2}}$
${y}^{\prime }={\left(1+\mathrm{cos}x\right)}^{1-2}$
Since, $\frac{{x}^{a}}{{x}^{b}}={x}^{a-b}$
${y}^{\prime }={\left(1+\mathrm{cos}x\right)}^{-1}$
${y}^{\prime }=\frac{1}{1+\mathrm{cos}x}$