Question

Find the derivatives w.r.t.x:
$\rm \frac{4x+5\sin x}{3x+7\cos x}$

Answer 1

here is ur Answer ✍️✍️

Answer 2

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

$\frac{d}{dx} \bigg(\frac{U}{V}\bigg ) = \frac{V \frac{dU}{dx} - U \frac{dV}{dx} }{ {V}^{2} } \\ \\ here \: U = 4x + 5sinx \\ \\ V = 3x + 7cos \: x$
$\frac{d}{dx} \bigg(\frac{4x + 5sin \: x}{3x + 7 \: cos \: x} \bigg) = \frac{(3x + 7 \: cos \: x) \frac{d(4x + 5sin \: x)}{dx} -( 4x + 5sin \: x )\frac{d(3x + 7 \: cos \: x)}{dx} }{( {3x + 7 \: cos \: x})^{2} } \\ \\ = \frac{(3x + 7 \: cos \: x) (4+ 5 \: cos \: x) -( 4x + 5sin \: x )(3 - 7 \: sin \: x) }{( {3x + 7 \: cos \: x})^{2} } \\ \\ = \frac{(12x + 15 \: x \: cos \: x + 28 \: cos \: x + 35 {cos}^{2}x - 12x+ 28x \: sin \: x - 15sin \: x + 35 {sin}^{2} x) }{( {3x + 7 \: cos \: x})^{2} } \\ \\\frac{d}{dx} \bigg(\frac{4x + 5sin \: x}{3x + 7 \: cos \: x} \bigg) = \frac{(15 \: x \: cos \: x + 28 \: cos \: x + 28x \: sin \: x - 15sin \: x + 35) }{( {3x + 7 \: cos \: x})^{2} }$
Hope it helps you.