University:
California State University, Northridge
Course:
MATH 351 | Differential Equations
Academic year:

2024
Views:

201

Pages:

3
Author:

Javon Neal

Question 1 Find the derivatives of 𝑔 ( 𝑧 ) = − 2 csc ⁡ 7 𝑧 + 8 sin ⁡ 3 𝑧 + 12 tan ⁡ 2 𝑧 g(z)=−2csc7z+8sin3z+12tan2z. Answer Using the sum rule of derivatives: 𝑑 𝑑 𝑧 [ 𝑔 ( 𝑧 ) + ℎ ( 𝑧 ) + 𝑘 ( 𝑧 ) ] = 𝑑 𝑑 𝑧 [ 𝑔 ( 𝑧 ) ] + 𝑑 𝑑 𝑧 [ ℎ ( 𝑧 ) ] + 𝑑 𝑑 𝑧 [ 𝑘 ( 𝑧 ) ] . dz d [g(z)+h(z)+k(z)]= dz d [g(z)]+ dz d [h(z)]+ dz d [k(z)]. Differentiate each term: For − 2 csc ⁡ 7 𝑧 −2csc7z: 𝑑 𝑑 𝑧 [ − 2 csc ⁡ 7 𝑧 ] = − 2 ( − 7 cot ⁡ 7 𝑧 csc ⁡ 7 𝑧 ) = 14 cot ⁡ 7 𝑧 csc ⁡ 7 𝑧 . dz d [−2csc7z]=−2(−7cot7zcsc7z)=14cot7zcsc7z. For 8 sin ⁡ 3 𝑧 8sin3z: 𝑑 𝑑 𝑧 [ 8 sin ⁡ 3 𝑧 ] = 8 ( 3 cos ⁡ 3 𝑧 ) = 24 cos ⁡ 3 𝑧 . dz d [8sin3z]=8(3cos3z)=24cos3z. For 12 tan ⁡ 2 𝑧 12tan2z: 𝑑 𝑑 𝑧 [ 12 tan ⁡ 2 𝑧 ] = 12 ( 2 sec ⁡ 2 2 𝑧 ) = 24 sec ⁡ 2 2 𝑧 . dz d [12tan2z]=12(2sec 2 2z)=24sec 2 2z. Combine the results: 𝑔 ′ ( 𝑧 ) = 14 cot ⁡ 7 𝑧 csc ⁡ 7 𝑧 + 24 cos ⁡ 3 𝑧 + 24 sec ⁡ 2 2 𝑧 . g ′ (z)=14cot7zcsc7z+24cos3z+24sec 2 2z. Question 2 Calculate the derivatives of 𝑓 ( 𝑥 ) = 4 − 𝑥 2 f(x)=4−x 2 and evaluate 𝑓 ′ ( − 3 ) , 𝑓 ′ ( 0 ) , 𝑓 ′ ( 1 ) f ′ (−3),f ′ (0),f ′ (1). Answer Differentiate 𝑓 ( 𝑥 ) = 4 − 𝑥 2 f(x)=4−x 2 : 𝑓 ′ ( 𝑥 ) = − 2 𝑥 . f ′ (x)=−2x. Evaluate at given points: At 𝑥 = − 3 x=−3: 𝑓 ′ ( − 3 ) = − 2 ( − 3 ) = 6. f ′ (−3)=−2(−3)=6. At 𝑥 = 0 x=0: 𝑓 ′ ( 0 ) = − 2 ( 0 ) = 0. f ′ (0)=−2(0)=0. At 𝑥 = 1 x=1: 𝑓 ′ ( 1 ) = − 2 ( 1 ) = − 2. f ′ (1)=−2(1)=−2. Question 3 Find 𝑔 ( 11 ) ( 𝑥 ) g (11) (x) for 𝑔 ( 𝑥 ) = 𝑒 7 𝑥 g(x)=e 7x . Answer Differentiate 𝑔 ( 𝑥 ) = 𝑒 7 𝑥 g(x)=e 7x : 𝑔 ′ ( 𝑥 ) = 7 𝑒 7 𝑥 . g ′ (x)=7e 7x . To find higher derivatives, observe the pattern: 𝑔 ′ ′ ( 𝑥 ) = 7 2 𝑒 7 𝑥 , 𝑔 ( 𝑛 ) ( 𝑥 ) = 7 𝑛 𝑒 7 𝑥 . g ′′ (x)=7 2 e 7x ,g (n) (x)=7 n e 7x . For 𝑛 = 11 n=11: 𝑔 ( 11 ) ( 𝑥 ) = 7 11 𝑒 7 𝑥 . g (11) (x)=7 11 e 7x . Question 4 Find the first partial derivatives of 𝑧 = ln ⁡ ( 𝑥 2 + 𝑦 2 ) z=ln(x 2 +y 2 ). Answer Partial derivative with respect to 𝑥 x: 𝑧 𝑥 = ∂ ∂ 𝑥 [ ln ⁡ ( 𝑥 2 + 𝑦 2 ) ] = 1 𝑥 2 + 𝑦 2 ⋅ ∂ ∂ 𝑥 ( 𝑥 2 + 𝑦 2 ) = 1 𝑥 2 + 𝑦 2 ⋅ 2 𝑥 . z x = ∂x ∂ [ln(x 2 +y 2 )]= x 2 +y 2 1 ⋅ ∂x ∂ (x 2 +y 2 )= x 2 +y 2 1 ⋅2x. Simplify: 𝑧 𝑥 = 2 𝑥 𝑥 2 + 𝑦 2 . z x = x 2 +y 2 2x . Partial derivative with respect to 𝑦 y: 𝑧 𝑦 = ∂ ∂ 𝑦 [ ln ⁡ ( 𝑥 2 + 𝑦 2 ) ] = 1 𝑥 2 + 𝑦 2 ⋅ ∂ ∂ 𝑦 ( 𝑥 2 + 𝑦 2 ) = 1 𝑥 2 + 𝑦 2 ⋅ 2 𝑦 . z y = ∂y ∂ [ln(x 2 +y 2 )]= x 2 +y 2 1 ⋅ ∂y ∂ (x 2 +y 2 )= x 2 +y 2 1 ⋅2y. Simplify: 𝑧 𝑦 = 2 𝑦 𝑥 2 + 𝑦 2 . z y = x 2 +y 2 2y . Question 5 Find the derivative of 𝑓 ( 𝑥 ) = ( 5 𝑥 2 − 2 𝑥 + 1 ) 3 f(x)=(5x 2 −2x+1) 3 . Answer Use the chain rule: 𝑑 𝑑 𝑥 [ 𝑢 𝑛 ] = 𝑛 𝑢 𝑛 − 1 𝑑 𝑢 𝑑 𝑥 , where 𝑢 = 5 𝑥 2 − 2 𝑥 + 1 . dx d [u n ]=nu n−1 dx du ,where u=5x 2 −2x+1. Differentiate: 𝑓 ′ ( 𝑥 ) = 3 ( 5 𝑥 2 − 2 𝑥 + 1 ) 2 ⋅ 𝑑 𝑑 𝑥 [ 5 𝑥 2 − 2 𝑥 + 1 ] . f ′ (x)=3(5x 2 −2x+1) 2 ⋅ dx d [5x 2 −2x+1]. Differentiate 5 𝑥 2 − 2 𝑥 + 1 5x 2 −2x+1: 𝑑 𝑑 𝑥 [ 5 𝑥 2 − 2 𝑥 + 1 ] = 10 𝑥 − 2. dx d [5x 2 −2x+1]=10x−2. Combine the results: 𝑓 ′ ( 𝑥 ) = 3 ( 5 𝑥 2 − 2 𝑥 + 1 ) 2 ( 10 𝑥 − 2 ) . f ′ (x)=3(5x 2 −2x+1) 2 (10x−2).