lim
𝑥→0sin𝑥
𝑥= 1
lim
𝑥→01−cos𝑥
𝑥= 0
lim
𝑥→01−cos𝑥
𝑥2=1
2
lim
𝑥→∞ (1+1
𝑥)𝑥=𝑒
lim
𝑥→0(1+ 𝑥)1
𝑥= 𝑒
lim
𝑥→0ln(1+𝑥)
𝑥= 1
lim
𝑥→0log𝑎(𝑥 + 1)
𝑥= log𝑎𝑒
lim
𝑥→0𝑒𝑥−1
𝑥= 1
lim
𝑥→0𝑎𝑥−1
𝑥=ln𝑎
lim
𝑥→0(1+𝑥)𝑘− 1
𝑥= 𝑘
𝑓′(𝑥)= lim
∆𝑥→0∆𝑦
∆𝑥 =lim
ℎ→0 𝑓(𝑐+ ℎ)−𝑓(𝑐)
ℎ
𝑦− 𝑓(𝑥0)= 𝑓′(𝑥0)(𝑥−𝑓(𝑥0))
𝑦 = 𝑘
𝑦′=0
𝑦 = 𝑎𝑟𝑐 tan𝑥
𝑦′=1
1+𝑥2
𝑦 = 𝑥
𝑦′=1
𝑦 = 𝑎𝑟𝑐 cot𝑥
𝑦′=− 1
1+𝑥2
𝑦 = 𝑥𝛼
𝑦′=𝛼𝑥𝛼−1
𝑦 = 𝑎𝑟𝑐 sin𝑥
𝑦′=1
√1−𝑥2
𝑦 = √𝑥
𝑦′=1
2√𝑥
𝑦 = 𝑎𝑟𝑐 cos𝑥
𝑦′=− 1
√1−𝑥2
𝑦 = 𝑎𝑥
𝑦′=𝑎𝑥ln𝑎
𝑦 = 𝑘 ∙ 𝑓(𝑥)
𝑦′=𝑘 ∙𝑓′(𝑥)
𝑦 = 𝑒𝑥
𝑦′=𝑒𝑥
𝑦 = 𝑓(𝑥)+𝑔(𝑥)
𝑦′=𝑓′(𝑥)+𝑔′(𝑥)
𝑦 = log𝑎𝑥
𝑦′=1
𝑥log𝑎𝑒
𝑦 = 𝑓(𝑥) ∙ 𝑔(𝑥)
𝑦′=𝑓′(𝑥)∙𝑔(𝑥)+ 𝑓(𝑥)∙ 𝑔′(𝑥)
𝑦 = ln𝑥
𝑦′=1
𝑥
𝑦 = 1
𝑓(𝑥)
𝑦′= − 𝑓′(𝑥)
𝑓′′(𝑥)
𝑦 = sin 𝑥
𝑦′=cos𝑥
𝑦 = 𝑓(𝑥)
𝑔(𝑥)
𝑦′=𝑓′(𝑥)∙𝑔(𝑥)−𝑓(𝑥) ∙𝑔′(𝑥)
𝑔2(𝑥)
𝑦 = cos 𝑥
𝑦′=−sin𝑥
𝑦 = 𝑓(𝑔(𝑥))
𝑦′=𝑓′(𝑔(𝑥)) ∙𝑔′(𝑥)
𝑦 = tan 𝑥
𝑦′=1
cos 2𝑥= 1 + tan 2𝑥
𝑦 = [𝑓(𝑥)]𝛼
𝑦′=𝛼[𝑓(𝑥)]𝛼−1 ∙𝑓′(𝑥)
𝑦 = cot 𝑥
𝑦′=1
sin 2𝑥= −(1 + cot 2𝑥)
𝑦 = [𝑓−1(𝑦)]
𝑦′=1
𝑓′(𝑥), 𝑐𝑜𝑛 𝑥 = 𝑓−1(𝑦)
∫𝑥𝛼𝑑𝑥 =𝑥𝛼+1
𝛼+ 1 + 𝑐, 𝑐𝑜𝑛 𝛼 ≠ −1
∫[𝑓(𝑥)]𝛼𝑓′(𝑥)𝑑𝑥 =[𝑓(𝑥)]𝛼+1
𝛼+ 1 + 𝑐, 𝑐𝑜𝑛 𝛼 ≠ −1
∫1
𝑥𝑑𝑥 = ln|𝑥|+𝑐
∫𝑓′(𝑥)
𝑓(𝑥) 𝑑𝑥 =ln|𝑓(𝑥)|+𝑐
∫𝑒𝑥𝑑𝑥 = 𝑒 𝑥+𝑐
∫𝑓′(𝑥)𝑒𝑓(𝑥) 𝑑𝑥 = 𝑒 𝑓(𝑥) +𝑐
∫𝑎𝑥𝑑𝑥 =𝑎𝑥
ln𝑎+𝑐
∫𝑓′(𝑥)𝑎𝑓(𝑥) 𝑑𝑥 =𝑎𝑓(𝑥)
ln𝑎+ 𝑐
∫sin𝑥𝑑𝑥 = −cos𝑥 + 𝑐
∫𝑓′(𝑥)sin𝑓(𝑥)𝑑𝑥 = − cos𝑓(𝑥)+ 𝑐
∫cos𝑥𝑑𝑥= sin 𝑥 +𝑐
∫𝑓′(𝑥)cos𝑓(𝑥)𝑑𝑥 = sin 𝑓(𝑥)+ 𝑐
∫1
cos 2𝑥𝑑𝑥= tan 𝑥 +𝑐
∫𝑓′(𝑥)
cos 2𝑓(𝑥)𝑑𝑥 =tan𝑓(𝑥) +𝑐
∫1
sin 2𝑥𝑑𝑥 = −cot 𝑥 + 𝑐
∫𝑓′(𝑥)
sin 2𝑓(𝑥)𝑑𝑥 = − cot 𝑓(𝑥)+𝑐
∫1
√1−𝑥2𝑑𝑥 =𝑎𝑟𝑐sin𝑥 + 𝑐
∫𝑓′(𝑥)
√1−[𝑓(𝑥)]2𝑑𝑥 = 𝑎𝑟𝑐 sin𝑓(𝑥) +𝑐
∫1
√1+𝑥2𝑑𝑥 =𝑎𝑟𝑐tan𝑥 + 𝑐
∫𝑓′(𝑥)
1+[𝑓(𝑥)]2𝑑𝑥= 𝑎𝑟𝑐 tan𝑓(𝑥) +𝑐
∫ln𝑥𝑑𝑥 = 𝑥ln𝑥 −𝑥 +𝑐
∫tan𝑥𝑑𝑥 =−ln|cos𝑥|+𝑐
∫cot𝑥𝑑𝑥=ln|sin𝑥|+ 𝑐
𝑓(𝑥)𝑔(𝑥) = 𝑒ln𝑓(𝑥)𝑔(𝑥) = 𝑒𝑔(𝑥)ln𝑓(𝑥)
