Question

if y= 2cos (x/2), then d^2y/dx^2=
don't understand how to solve this :(

Answer 1

\[y=2 \cos \frac{ x }{ 2 }\]
\[\frac{ dy }{ dx }=-2 \sin \frac{ x }{ 2 }*\frac{ 1 }{ 2 }=-\sin \frac{ x }{ 2 }\]
\[\frac{ d^2y }{ dx^2 }=-\cos \frac{ x }{ 2 }*\frac{ d }{ dx }\left( \frac{ x }{ 2 } \right)=?\]

Answer 2

uhm im not quite sure i understand

Answer 3

\[\frac{ d }{ dx }\left( \cos \frac{ x }{ 2 } \right)=\frac{ d }{ dx }\left( \cos... \right)\frac{ d }{ dx }\left( \frac{ x }{ 2 } \right)\]