Question

Find the derivatives w.r.t.x:
$\rm \frac{2\sin x}{1+3\cos x}$

Answer 1

HOPE IT HELPS U ✌️✌️

Answer 2

We know that such form of expression can solve by U/V method of differentiation

$\frac{d}{dx} \bigg(\frac{U}{V}\bigg ) = \frac{V \frac{dU}{dx} - U \frac{dV}{dx} }{ {V}^{2} } \\ \\ here \: U = 2sinx \\ \\ V = 1 + 3cos \: x$
$\frac{d}{dx} \bigg(\frac{2sin \: x}{1+ 3\: cos \: x} \bigg) = \frac{(1+3 \: cos \: x) \frac{d(2sin \: x)}{dx} -( 2sin \: x )\frac{d(1+3 \: cos \: x)}{dx} }{( {1+3 \: cos \: x})^{2} }$

$\frac{d}{dx} \bigg(\frac{2sin \: x}{1+ 3\: cos \: x} \bigg) = \frac{(1+3 \: cos \: x)(2 \: cos \: x) -( 2sin \: x )(0 - 3 \: sin \: x) }{( {1+3 \: cos \: x})^{2} } \\ \\ = \frac{(2 \: cos \: x+6 \: {cos}^{2}x + 6 {sin}^{2} x)}{( {1+3 \: cos \: x})^{2} } \\ \\ we \: know \: that \: {cos}^{2}x + {sin}^{2} x = 1 \\ \\ \frac{d}{dx} \bigg(\frac{2sin \: x}{1+ 3\: cos \: x} \bigg)= \frac{(2 \: cos \: x+6)}{( {1+3 \: cos \: x})^{2} }$

Hope it helps you.