Question

Differentciation with chain rule: log(sin x/1+cos x) ​

Answer 1

$</p><p></p><p>\sf \rightarrow \quad \frac{d}{dx} \bigg(log \:( \frac{sin \: x}{1 + cos \: x} )\bigg)</p><p> \\ \\ \sf \rightarrow \quad \frac{d}{dx} \bigg \{ \frac{ln \big( \frac{sin \: x}{ 1 + cos \: x} \big)}{ln \: 10} \bigg \} \\ \\ \sf \rightarrow \quad \frac{1}{ln \: 10} \bigg\{ \frac{d}{dx} \bigg(ln \: \frac{sin \: x}{1 + cos \: x} \bigg) \bigg \} \\ \\ \sf \rightarrow \quad \frac{1}{ln \: 10} \bigg \{ \frac{d \bigg(ln \: \frac{sin \: x}{1 + cos \: x} \bigg)}{d \frac{sin \: x}{1 + cos \: x} } \cdot \frac{d \big( \frac{sin \: x}{1 + cos \: x} \big)}{dx} \bigg \} \\ \\ \sf \rightarrow \quad \frac{1}{ln \: 10} \cdot \frac{1 + cos \: x}{sin \: x} \cdot \bigg \{ \frac{ \frac{d(sin \: x)}{dx} \cdot(1 + cos \: x) - \frac{d(1 + cos \: x)}{dx} \cdot sin \: x }{ {(1 + cos \: x)}^{2} } \bigg \} \\ \\ \sf \rightarrow \quad \frac{1}{ln \: 10} \cdot \frac{1 + cos \: x}{sin \: x} \bigg( \frac{ cos \: x(1 + cos \: x) - ( - sin \: x \cdot sin \: x) }{ {(1 + cos \: x)}^{2} } \bigg) \\ \\ \sf \rightarrow \quad \frac{1}{ln \: 10} \cdot \frac{1 + cos \: x}{sin \: x} \bigg( \frac{cos \: x \: + {cos}^{2} \: x + {sin}^{2} \: x}{ {(1 + cos \: x)}^{2} } \bigg) \\ \\ \sf \rightarrow \quad \frac{1}{ln \: 10} \cdot \frac{ \cancel{1 + cos \: x}}{sin \: x} \cdot \frac{ \cancel{1 + cos \: x}}{ \cancel{(1 + cos \: x)} \cancel{(1 + cos \: x)} } \\ \\ \sf \rightarrow \quad \frac{1}{ln \: 10} \cdot \frac{1}{sin \: x} \\ \\ \sf \rightarrow \quad \frac{cosec \: x}{ln \: 10} </p><p>$