yusufca35 14:02 09 Nis 2012 #1
Alp50 16:10 09 Nis 2012 #2
1)f'(x)=8x³−6x+3
f2(−2)=−64+12+3=−49
2)y'=−2sin2x.cos(cos2x)−[−2cos2x.sin(sin2x)]
y'=−2sin2x.cos(cos2x)+2cos2x.sin(sin2x)
x=pi/4 ise y'=−2.1+0=−2
3)3x²−2=1
3(x−1)(x+1)=0
x=1 ve x=−1
yerine koyarsak (1,5) ve (−1,7) noktaları
4)x⁴−2x²+3x+c
5)sinx=u cosxdx=du
integralin içi √udu oldu
integralini alınca 2/3.sinx√sinx+c