exercice dérivation 2

salut



bonne chance à tous

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1)
li yakon x 3onsore mina l majal ]0.pi[ /x youkhalif pi/2
ladayna [f(x)-f(pi/2)]/x-pi/2=[lcosxlv1-cosx]/x-pi/2
ida kana 0<x<pi/2 fa ina cosx>0 wa minho
[f(x)-f(pi/2)]/x-pi/2=cosx/x-pi/2*v1-cosx
ida kana pi/2<x<pi fa ina cosx<0 waminho
[f(x)-f(pi/2)]/x-pi/2=-cosx/x-pi/2*v1-cosx
ladayna lim v1-cosx =v1-cospi/2=1 (x vers pi/2 )
donc lim cosx/x-pi/2=lim cosx-cos(pi/2)/x-pi/2=cos'(pi/2)
donc lim cosx/x-pi/2=-sin(pi/2)=-1
wamiho nastantij anaho
3indama yakono x<pi/2
lim f(x)-f(pi/2)/x-pi/2=lim cosx/x-pi/2*v1-cosx=-1
wa 3indama yakono x>pi/2
lim f(x)-f(pi/2)/x-pi/2=lim -cosx/x-pi/2*v1-cosx=1
wabitali fa ina dala f(x) 9abila lil ichti9a9 3ala lyamin wa 3ala lyasare fi pi/2 walakinaha rayre 9abila lil ichti9a9 fi pi/2

2)
g(x)=(f(x))²
g(x)=lcosxl²(1-cosx)
g(x)=cos²x(1-cosx)
g(x)=cos²x-cos^3x
wa bima ana dala cosx 9abila lil ichti9a9 3ala IR fa ina dala g(x) 9abila lil ichti9a9 3ala l majal ]0,pi[
donc ladayna
likoli x min ]0,pi[
g'(x)=2cosxcos'x-3cos²xcos'x
g'(x)=-2cosxsinx+3cos²xsinx
g'(x)=3cos²xsinx-sin2x

3)
likoli x min ]0,pi[
f(x)=vg(x)
donc kml mn 3andak
tajid ana likoli x min]0,pi[-pi/2
f'(x)=3cos²xsinx-sin2x/3lcoxlv1-cosx
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