Derivatives #26 - Questions and Answers

Question 1 Use the limit definition of partial derivatives to evaluate 𝑓 𝑥 ( 𝑥 , 𝑦 ) f x ​ (x,y) and 𝑓 𝑦 ( 𝑥 , 𝑦 ) f y ​ (x,y) for the following function: 𝑓 ( 𝑥 , 𝑦 ) = 𝑥 𝑦 f(x,y)= y x ​ . Answer Step 1 The given function is 𝑓 ( 𝑥 , 𝑦 ) = 𝑥 𝑦 f(x,y)= y x ​ . Step 2 Partial derivative with respect to 𝑥 x: 𝑓 𝑥 ( 𝑥 , 𝑦 ) = lim ⁡ ℎ → 0 𝑓 ( 𝑥 + ℎ , 𝑦 ) − 𝑓 ( 𝑥 , 𝑦 ) ℎ . f x ​ (x,y)= h→0 lim ​ h f(x+h,y)−f(x,y) ​ . Substitute: 𝑓 𝑥 ( 𝑥 , 𝑦 ) = lim ⁡ ℎ → 0 𝑥 + ℎ 𝑦 − 𝑥 𝑦 ℎ . f x ​ (x,y)= h→0 lim ​ h y x+h ​ − y x ​ ​ . Simplify: 𝑓 𝑥 ( 𝑥 , 𝑦 ) = lim ⁡ ℎ → 0 𝑥 + ℎ − 𝑥 𝑦 ℎ = lim ⁡ ℎ → 0 ℎ 𝑦 ℎ = 1 𝑦 . f x ​ (x,y)= h→0 lim ​ h y x+h−x ​ ​ = h→0 lim ​ h y h ​ ​ = y 1 ​ . Partial derivative with respect to 𝑦 y: 𝑓 𝑦 ( 𝑥 , 𝑦 ) = lim ⁡ ℎ → 0 𝑓 ( 𝑥 , 𝑦 + ℎ ) − 𝑓 ( 𝑥 , 𝑦 ) ℎ . f y ​ (x,y)= h→0 lim ​ h f(x,y+h)−f(x,y) ​ . Substitute: 𝑓 𝑦 ( 𝑥 , 𝑦 ) = lim ⁡ ℎ → 0 𝑥 𝑦 + ℎ − 𝑥 𝑦 ℎ . f y ​ (x,y)= h→0 lim ​ h y+h x ​ − y x ​ ​ . Simplify using L'Hôpital's rule: 𝑓 𝑦 ( 𝑥 , 𝑦 ) = − 𝑥 𝑦 2 . f y ​ (x,y)=− y 2 x ​ . Final Results: 𝑓 𝑥 ( 𝑥 , 𝑦 ) = 1 𝑦 , 𝑓 𝑦 ( 𝑥 , 𝑦 ) = − 𝑥 𝑦 2 . f x ​ (x,y)= y 1 ​ ,f y ​ (x,y)=− y 2 x ​ . Question 2 Define Higher Order Derivatives. Answer Step 1 Higher order derivatives refer to derivatives taken multiple times successively. Step 2 The first derivative of a function 𝑓 ( 𝑥 ) f(x) is 𝑓 ′ ( 𝑥 ) f ′ (x). The second derivative of 𝑓 ( 𝑥 ) f(x) is 𝑓 ′ ′ ( 𝑥 ) f ′′ (x), the derivative of 𝑓 ′ ( 𝑥 ) f ′ (x). Higher order derivatives include third, fourth, and successive derivatives, written as 𝑓 ( 𝑛 ) ( 𝑥 ) f (n) (x), where 𝑛 n represents the order. Example: The 𝑛 n-th order derivative of 𝑓 ( 𝑥 ) f(x) is written as 𝑓 ( 𝑛 ) ( 𝑥 ) f (n) (x). Question 3 Evaluate the derivative of 𝑥 tan ⁡ 𝑥 x tanx . Answer Step 1 Let 𝑦 = 𝑥 tan ⁡ 𝑥 y=x tanx . Taking ln ⁡ ln on both sides: ln ⁡ 𝑦 = ( tan ⁡ 𝑥 ) ln ⁡ 𝑥 . lny=(tanx)lnx. Step 2 Differentiate both sides with respect to 𝑥 x: 1 𝑦 𝑑 𝑦 𝑑 𝑥 = tan ⁡ 𝑥 ⋅ 1 𝑥 + ln ⁡ 𝑥 ⋅ sec ⁡ 2 𝑥 . y 1 ​ dx dy ​ =tanx⋅ x 1 ​ +lnx⋅sec 2 x. Simplify: 𝑑 𝑦 𝑑 𝑥 = 𝑦 ( tan ⁡ 𝑥 𝑥 + ln ⁡ 𝑥 ⋅ sec ⁡ 2 𝑥 ) . dx dy ​ =y( x tanx ​ +lnx⋅sec 2 x). Substitute 𝑦 = 𝑥 tan ⁡ 𝑥 y=x tanx : 𝑑 𝑦 𝑑 𝑥 = 𝑥 tan ⁡ 𝑥 ( tan ⁡ 𝑥 𝑥 + ln ⁡ 𝑥 ⋅ sec ⁡ 2 𝑥 ) . dx dy ​ =x tanx ( x tanx ​ +lnx⋅sec 2 x). Question 4 Evaluate the derivative of ln ⁡ 3 ( 3 𝑥 2 + 2 ) ln 3 (3x 2 +2). Answer 1 Using the chain rule: 𝑑 𝑑 𝑥 ( ln ⁡ 3 ( 3 𝑥 2 + 2 ) ) = 3 ln ⁡ 2 ( 3 𝑥 2 + 2 ) ⋅ 1 3 𝑥 2 + 2 ⋅ 6 𝑥 . dx d ​ (ln 3 (3x 2 +2))=3ln 2 (3x 2 +2)⋅ 3x 2 +2 1 ​ ⋅6x. Simplify: 𝑑 𝑑 𝑥 ( ln ⁡ 3 ( 3 𝑥 2 + 2 ) ) = 18 𝑥 ln ⁡ 2 ( 3 𝑥 2 + 2 ) 3 𝑥 2 + 2 . dx d ​ (ln 3 (3x 2 +2))= 3x 2 +2 18xln 2 (3x 2 +2) ​ . Answer 2 Alternatively, rewrite the derivative as: 𝑑 𝑑 𝑥 ( ln ⁡ 3 ( 3 𝑥 2 + 2 ) ) = 3 ln ⁡ 2 ( 3 𝑥 2 + 2 ) ⋅ 6 𝑥 3 𝑥 2 + 2 . dx d ​ (ln 3 (3x 2 +2))=3ln 2 (3x 2 +2)⋅ 3x 2 +2 6x ​ . Result: 18 𝑥 ln ⁡ 2 ( 3 𝑥 2 + 2 ) 3 𝑥 2 + 2 . 3x 2 +2 18xln 2 (3x 2 +2) ​ . Question 5 Derivative of the product of trigonometric functions: 3 sin ⁡ ( 𝑥 ) cot ⁡ ( 𝑥 ) sin ⁡ 2 ( 𝑥 ) sin 2 (x) 3sin(x)cot(x) ​ . Answer We need to differentiate 3 sin ⁡ ( 𝑥 ) cot ⁡ ( 𝑥 ) sin ⁡ 2 ( 𝑥 ) sin 2 (x) 3sin(x)cot(x) ​ . Use the product rule: 𝑑 𝑑 𝑥 [ 3 sin ⁡ ( 𝑥 ) cot ⁡ ( 𝑥 ) ] = 3 [ cot ⁡ ( 𝑥 ) ⋅ cos ⁡ ( 𝑥 ) − sin ⁡ ( 𝑥 ) csc ⁡ 2 ( 𝑥 ) ] . dx d ​ [3sin(x)cot(x)]=3[cot(x)⋅cos(x)−sin(x)csc 2 (x)]. Simplify using trigonometric identities: cos ⁡ 2 ( 𝑥 ) − sin ⁡ 2 ( 𝑥 ) sin ⁡ 3 ( 𝑥 ) = − 3 sin ⁡ ( 𝑥 ) sin ⁡ 2 ( 𝑥 ) = − 3 cot ⁡ ( 𝑥 ) . sin 3 (x) cos 2 (x)−sin 2 (x) ​ =− sin 2 (x) 3sin(x) ​ =−3cot(x). Result: The derivative simplifies to − 3 sin ⁡ ( 𝑥 ) −3sin(x).