Question

Differentiate

Answer 1

woops wrong equation it should be this one

Answer 2

\[y=\frac{ u }{ v }\]\[\frac{ dy }{ dx }=\frac{ v \frac{ du }{ dx }-u \frac{ dv }{ dx} }{ v^2 }\]
or\[y=\frac{ 1+\sin x }{ 1-\cos x }=\frac{ \cos ^2\frac{ x }{ 2 }+\sin ^2\frac{ x }{ 2 }+2\sin \frac{ x }{ 2 }\cos \frac{ x }{ 2 } }{ 2\sin ^2\frac{ x }{ 2 } }\]
\[=\frac{ 1 }{ 2 }\cot ^2\frac{ x }{ 2 }+\frac{ 1 }{ 2 }+\cot \frac{ x }{ 2 }\]
find \[\frac{ dy }{ dx }\]

Answer 3

im getting (-sin(x)+cos(x)-1)/((1-cos(x))^2)

Answer 4

\[\frac{ dy }{ dx }=\frac{\left( 1-\cos x \right)\cos x-\left( 1+\sin x \right)\sin x }{ \left( 1-\cos x \right)^2 }\]
\[=\frac{ \cos x-\cos ^2x-\sin x-\sin ^2x}{ \left( 1-\cos x \right)^2 }\]
\[=\frac{ \cos x-\sin x-1 }{ \left( 1-\cos x \right)^2 }\]
you are correct