derivative of y=ln square root (1+sinx/1-sinx)
ok let \[u = \frac{ 1+\sin(x) }{ 1-\sin(x) }\]
use the Quotient rule\[\frac{ vdu-udv }{ v^2 }\]
Better be specific though... \[\huge \frac{d}{dx}\frac{f(x)}{g(x)}=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}\]
\[\frac{ (1-\sin(x))(\cos(x))-(1+\sin(x))(\cos(x)) }{ (1-\sin(x))^2 }\]
simplify the above expression
\[y =\frac{ 1 }{ 2 }\ln(u)\]
\[\frac{ dy }{ du }=\frac{ 1 }{ 2 }\frac{ 1 }{ u }\]
apply the chain rule
\[\frac{ dy }{ dx }=\frac{ dy }{ du}\frac{ du }{ dx }\]
@mathsmind Something's not quite right... I think it should be \[\large\frac{ (1-\sin(x))(\cos(x))-(1+\sin(x))(-\cos(x)) }{ (1-\sin(x))^2 }\]
yes the minus sign i deleted by mistake i just noticed
final answer is
\[\frac{ 1 }{ 2 }\frac{ \ln(\frac{ \cos(x) }{ 1-\sin(x) }+\frac{ (1+\sin(x))\cos(x) }{ (1-\sin(x))^2 }) }{ \sqrt{\frac{ 1+\sin(x) }{ 1-\sin(x) }} }\]
recall that \[\frac{ d }{ dx }\sin(x) = \cos(x)\]
A bit hard to read, @mathsmind \[\huge \frac{ 1 }{ 2 }\frac{ \ln(\frac{ \cos(x) }{ 1-\sin(x) }+\frac{ (1+\sin(x))\cos(x) }{ (1-\sin(x))^2 }) }{ \sqrt{\frac{ 1+\sin(x) }{ 1-\sin(x) }} }\]
\[\frac{ d }{ dx}(1+\sin(x))=\cos(x)\]
change the resolution of the desktop or use opera browser
or just place \large or \huge before the entire thingy... :)
hehehe whatever
You couldn't be bothered to type an extra five characters? -.-
i am trying to get used to latex, so all my mind is focusing on coding
i need to memorize the user manual better
your resolution is too low that's why you are getting big fonts, at least set windows to small fonts
from the control pannel
Too much trouble :D
hehehe ok
i can see the full equation
acids taste sour and bases taste bitter
Where does this come from? 0.o
how did this post come into mathematics? errrr
Join our real-time social learning platform and learn together with your friends!