Exercicios Calculo (133)

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147 
 
𝑸𝒖𝒆𝒔𝒕ã𝒐 𝟏. 
 
𝑎) 𝑆𝑒 𝑓 𝑓𝑜𝑟 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖á𝑣𝑒𝑙 𝑒 𝑔(𝑥) = 𝑥𝑓(𝑥),𝑢𝑠𝑒 𝑎 𝑑𝑒𝑓𝑖𝑛𝑖çã𝑜 𝑑𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 
𝑝𝑎𝑟𝑎 𝑚𝑜𝑠𝑡𝑟𝑎𝑟 𝑞𝑢𝑒 𝑔′(𝑥) = 𝑥𝑓 ′(𝑥) + 𝑓(𝑥). 
 
𝐶𝑜𝑚𝑜 𝑓 é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖á𝑣𝑒𝑙, 𝑒𝑛𝑡ã𝑜 
 
𝑓 ′(𝑥) = lim
∆𝑥→0
𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)
∆𝑥
 
 
𝑃𝑒𝑙𝑎 𝑑𝑒𝑓𝑖𝑛𝑖çã𝑜 𝑑𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎, 
 
𝑔′(𝑥) = lim
∆𝑥→0
𝑔(𝑥 + ∆𝑥) − 𝑔(𝑥)
∆𝑥
 
 = lim
∆𝑥→0
(𝑥 + ∆𝑥)𝑓(𝑥 + ∆𝑥) − 𝑥𝑓(𝑥)
∆𝑥
 
 = lim
∆𝑥→0
𝑥[𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)]+ ∆𝑥𝑓(𝑥 + ∆𝑥)
∆𝑥
 
 = lim
∆𝑥→0
[
𝑥. [𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)]
∆𝑥
+
∆𝑥.𝑓(𝑥 + ∆𝑥)
∆𝑥
] 
 = 𝑥. lim
∆𝑥→0
𝑓(𝑥 + ∆𝑥)− 𝑓(𝑥)
∆𝑥
+ lim
∆𝑥→0
𝑓(𝑥 + ∆𝑥) ; 
 = 𝑥. 𝑓 ′(𝑥)+ 𝑓(𝑥) 
 
𝐿𝑜𝑔𝑜, 𝑔′(𝑥) = 𝑥𝑓 ′(𝑥) + 𝑓(𝑥). 
 
𝑏) 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑒 𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑑𝑒 𝑎𝑏𝑠𝑐𝑖𝑠𝑠𝑎𝑠 𝑎 𝑒 -𝑎 𝑑𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑎 𝑐𝑢𝑟𝑣𝑎 𝑦 = 𝑥2, 𝑜𝑛𝑑𝑒 𝑎 ≠ 0. 
𝐸𝑛𝑐𝑜𝑛𝑡𝑟𝑒 𝑎𝑠 𝑒𝑞𝑢𝑎çõ𝑒𝑠 𝑑𝑎𝑠 𝑟𝑒𝑡𝑎𝑠 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒𝑠 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑛𝑒𝑠𝑡𝑒𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑒 𝑒𝑚 𝑠𝑒𝑔𝑢𝑖𝑑𝑎 
𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑎 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑎𝑠 𝑟𝑒𝑡𝑎𝑠 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒𝑠 𝑛𝑒𝑠𝑡𝑒𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑠𝑒𝑗𝑎𝑚 𝑜𝑟𝑡𝑜𝑔𝑜𝑛𝑎𝑖𝑠. 
 
𝑃𝑜𝑛𝑡𝑜𝑠 𝐴(𝑎, 𝑎2) 𝑒 𝐵(−𝑎,𝑎2). 
 
𝑂 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑎 𝑐𝑢𝑟𝑣𝑎 𝑦 = 𝑥2 𝑛𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 
𝐴 𝑒 𝐵 𝑠ã𝑜 𝑛𝑢𝑚𝑒𝑟𝑖𝑐𝑎𝑚𝑒𝑛𝑡𝑒 𝑖𝑔𝑢𝑎𝑖𝑠 𝑎𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑑𝑎 𝑓𝑢𝑛çã𝑜 𝑑𝑎 𝑐𝑢𝑟𝑣𝑎 𝑛𝑎𝑠 
𝑎𝑏𝑠𝑐𝑖𝑠𝑠𝑎𝑠 𝑎 𝑒 -𝑎, 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑎𝑚𝑒𝑛𝑡𝑒 𝑚𝐴 𝑒 𝑚𝐵.𝐿𝑜𝑔𝑜, 
 
𝑦′ = 2𝑥 ⟹ 
𝑚𝐴 = 2𝑎
𝑚𝐵 = −2𝑎
 
 
𝐸𝑞𝑢𝑎çõ𝑒𝑠 𝑑𝑎𝑠 𝑟𝑒𝑡𝑎𝑠 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒𝑠 𝑛𝑒𝑠𝑡𝑒𝑠 𝑝𝑜𝑛𝑡𝑜𝑠: 
 
𝑁𝑜 𝑝𝑜𝑛𝑡𝑜 𝐴: 𝑦 − 𝑎2 = 2𝑎(𝑥 − 𝑎) ⟹ 𝑦 = 2𝑎𝑥 − 𝑎2 
𝑁𝑜 𝑝𝑜𝑛𝑡𝑜 𝐵: 𝑦− 𝑎2 = −2𝑎(𝑥 − (−𝑎))⟹ 𝑦 = −2𝑎𝑥 − 𝑎2. 
 
𝑃𝑎𝑟𝑎 𝑞𝑢𝑒 𝑒𝑠𝑡𝑎𝑠 𝑟𝑒𝑡𝑎𝑠 𝑠𝑒𝑗𝑎𝑚 𝑑𝑖𝑡𝑎𝑠 𝑜𝑟𝑡𝑜𝑔𝑜𝑛𝑎𝑖𝑠, 𝑜 𝑝𝑟𝑜𝑑𝑢𝑡𝑜 𝑑𝑒 𝑠𝑒𝑢𝑠 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠 
𝑎𝑛𝑔𝑢𝑙𝑎𝑟𝑒𝑠 𝑑𝑒𝑣𝑒 𝑠𝑒𝑟 𝑖𝑔𝑢𝑎𝑙 𝑎 − 1. 𝐶𝑜𝑚 𝑖𝑠𝑠𝑜 … 
 
𝑚𝐴. 𝑚𝐵 = −1 
(2𝑎). (−2𝑎) = −1