MAT1512 ASSIGNMENT 3 2021
Question 1
(a) 𝑓(𝑥) = ln(𝑥 + 𝑙𝑛𝑥)
Method 1 (Differentiate immediately)
1 𝑑
𝑓 ′ (𝑥) = × (𝑥 + 𝑙𝑛𝑥)
𝑥 + 𝑙𝑛𝑥 𝑑𝑥
1 1
𝑓 ′ (𝑥) = × (1 + )
𝑥 + 𝑙𝑛𝑥 𝑥
1 1 1
𝑓 ′ (𝑥) = ×( + )
𝑥 + 𝑙𝑛𝑥 1 𝑥
1 𝑥 1
𝑓 ′ (𝑥) = ×( + )
𝑥 + 𝑙𝑛𝑥 𝑥 𝑥
1 𝑥+1
𝑓 ′ (𝑥) = ×( )
𝑥 + 𝑙𝑛𝑥 𝑥
𝑥+1
𝑓 ′ (𝑥) =
𝑥(𝑥 + 𝑙𝑛𝑥)
𝑥+1
𝑓 ′ (𝑥) =
𝑥 2 + 𝑥𝑙𝑛𝑥
Method 2 (Implicit differentiation)
𝑓(𝑥) = ln(𝑥 + 𝑙𝑛𝑥)
𝑓 = ln(𝑥 + 𝑙𝑛𝑥)
𝑒 𝑓 = 𝑒 ln(𝑥+𝑙𝑛𝑥)
, 𝑒 𝑓 = 𝑥 + 𝑙𝑛𝑥
𝑑 𝑓 𝑑
𝑒 = (𝑥 + 𝑙𝑛𝑥)
𝑑𝑥 𝑑𝑥
𝑑𝑓 1
𝑒𝑓 × =1+
𝑑𝑥 𝑥
𝑑𝑓 1 1
= 𝑓 (1 + )
𝑑𝑥 𝑒 𝑥
𝑑𝑓 1 1
= ln(𝑥+𝑙𝑛𝑥) (1 + )
𝑑𝑥 𝑒 𝑥
𝑑𝑓 1 1
= (1 + )
𝑑𝑥 (𝑥 + 𝑙𝑛𝑥) 𝑥
1 1
𝑓 ′ (𝑥) = (1 + )
(𝑥 + 𝑙𝑛𝑥) 𝑥
1 𝑥 1
𝑓 ′ (𝑥) = ×( + )
(𝑥 + 𝑙𝑛𝑥) 𝑥 𝑥
1 𝑥+1
𝑓 ′ (𝑥) = ×( )
(𝑥 + 𝑙𝑛𝑥) 𝑥
𝑥+1
𝑓 ′ (𝑥) =
𝑥(𝑥 + 𝑙𝑛𝑥)
𝑥+1
𝑓 ′ (𝑥) =
𝑥2+ 𝑥𝑙𝑛𝑥
𝑥−1
(b) 𝑔(𝑥) = √𝑥 4 +1
Method 1 (Differentiate immediately)
𝑥−1
𝑔(𝑥) = √ 4
𝑥 +1
𝑥 − 1 1⁄2
𝑔(𝑥) = ( )
𝑥4 + 1
Question 1
(a) 𝑓(𝑥) = ln(𝑥 + 𝑙𝑛𝑥)
Method 1 (Differentiate immediately)
1 𝑑
𝑓 ′ (𝑥) = × (𝑥 + 𝑙𝑛𝑥)
𝑥 + 𝑙𝑛𝑥 𝑑𝑥
1 1
𝑓 ′ (𝑥) = × (1 + )
𝑥 + 𝑙𝑛𝑥 𝑥
1 1 1
𝑓 ′ (𝑥) = ×( + )
𝑥 + 𝑙𝑛𝑥 1 𝑥
1 𝑥 1
𝑓 ′ (𝑥) = ×( + )
𝑥 + 𝑙𝑛𝑥 𝑥 𝑥
1 𝑥+1
𝑓 ′ (𝑥) = ×( )
𝑥 + 𝑙𝑛𝑥 𝑥
𝑥+1
𝑓 ′ (𝑥) =
𝑥(𝑥 + 𝑙𝑛𝑥)
𝑥+1
𝑓 ′ (𝑥) =
𝑥 2 + 𝑥𝑙𝑛𝑥
Method 2 (Implicit differentiation)
𝑓(𝑥) = ln(𝑥 + 𝑙𝑛𝑥)
𝑓 = ln(𝑥 + 𝑙𝑛𝑥)
𝑒 𝑓 = 𝑒 ln(𝑥+𝑙𝑛𝑥)
, 𝑒 𝑓 = 𝑥 + 𝑙𝑛𝑥
𝑑 𝑓 𝑑
𝑒 = (𝑥 + 𝑙𝑛𝑥)
𝑑𝑥 𝑑𝑥
𝑑𝑓 1
𝑒𝑓 × =1+
𝑑𝑥 𝑥
𝑑𝑓 1 1
= 𝑓 (1 + )
𝑑𝑥 𝑒 𝑥
𝑑𝑓 1 1
= ln(𝑥+𝑙𝑛𝑥) (1 + )
𝑑𝑥 𝑒 𝑥
𝑑𝑓 1 1
= (1 + )
𝑑𝑥 (𝑥 + 𝑙𝑛𝑥) 𝑥
1 1
𝑓 ′ (𝑥) = (1 + )
(𝑥 + 𝑙𝑛𝑥) 𝑥
1 𝑥 1
𝑓 ′ (𝑥) = ×( + )
(𝑥 + 𝑙𝑛𝑥) 𝑥 𝑥
1 𝑥+1
𝑓 ′ (𝑥) = ×( )
(𝑥 + 𝑙𝑛𝑥) 𝑥
𝑥+1
𝑓 ′ (𝑥) =
𝑥(𝑥 + 𝑙𝑛𝑥)
𝑥+1
𝑓 ′ (𝑥) =
𝑥2+ 𝑥𝑙𝑛𝑥
𝑥−1
(b) 𝑔(𝑥) = √𝑥 4 +1
Method 1 (Differentiate immediately)
𝑥−1
𝑔(𝑥) = √ 4
𝑥 +1
𝑥 − 1 1⁄2
𝑔(𝑥) = ( )
𝑥4 + 1