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

2024
Views:

233

Pages:

4
Author:

Javon Neal

Question 1 First Partial Derivatives Given 𝑧 = 𝑓 ( 𝑥 ) 𝑔 ( 𝑦 ) z=f(x)g(y), find 𝑧 𝑥 + 𝑧 𝑦 z x +z y . Answer Step 1 Given function 𝑧 = 𝑓 ( 𝑥 ) 𝑔 ( 𝑦 ) z=f(x)g(y). When we find the partial derivative with respect to 𝑥 x, 𝑦 y is treated as a constant: ∂ 𝑧 ∂ 𝑥 = 𝑓 𝑥 ( 𝑥 ) 𝑔 ( 𝑦 ) = 𝑧 𝑥 ∂x ∂z =f x (x)g(y)=z x When we find the partial derivative with respect to 𝑦 y, 𝑥 x is treated as a constant: ∂ 𝑧 ∂ 𝑦 = 𝑓 ( 𝑥 ) 𝑔 𝑦 ( 𝑦 ) = 𝑧 𝑦 ∂y ∂z =f(x)g y (y)=z y Step 2 Partial derivatives: 𝑧 𝑥 = 𝑔 ( 𝑦 ) 𝑓 ′ ( 𝑥 ) and 𝑧 𝑦 = 𝑓 ( 𝑥 ) 𝑔 ′ ( 𝑦 ) z x =g(y)f ′ (x)andz y =f(x)g ′ (y) Hence: 𝑧 𝑥 + 𝑧 𝑦 = 𝑔 ( 𝑦 ) 𝑓 ′ ( 𝑥 ) + 𝑓 ( 𝑥 ) 𝑔 ′ ( 𝑦 ) z x +z y =g(y)f ′ (x)+f(x)g ′ (y) Question 2 Find the first partial derivatives of the function 𝑓 ( 𝑥 , 𝑡 ) = 𝑥 ln ⁡ 𝑡 f(x,t)= x lnt. Answer 1 Step 1: Find the partial derivative with respect to 𝑥 x (treat 𝑡 t as constant): 𝑓 𝑥 = 1 2 𝑥 ln ⁡ 𝑡 f x = 2 x 1 lnt Step 2: Find the partial derivative with respect to 𝑡 t (treat 𝑥 x as constant): 𝑓 𝑡 = 𝑥 ⋅ 1 𝑡 = 𝑥 𝑡 f t = x ⋅ t 1 = t x Answer 2 For 𝑥 x, since ln ⁡ 𝑡 lnt is treated as constant: ∂ 𝑓 ∂ 𝑥 = ∂ ∂ 𝑥 ( 𝑥 ln ⁡ 𝑡 ) = 1 2 𝑥 ln ⁡ 𝑡 ∂x ∂f = ∂x ∂ ( x lnt)= 2 x 1 lnt For 𝑡 t, since 𝑥 x is treated as constant: ∂ 𝑓 ∂ 𝑡 = 𝑥 ⋅ 1 𝑡 = 𝑥 𝑡 ∂t ∂f = x ⋅ t 1 = t x Thus: ∂ 𝑓 ∂ 𝑥 = 1 2 𝑥 ln ⁡ 𝑡 and ∂ 𝑓 ∂ 𝑡 = 𝑥 𝑡 ∂x ∂f = 2 x 1 lntand ∂t ∂f = t x Question 3 Find the first partial derivatives of the function 𝑓 ( 𝑥 , 𝑦 ) = 𝑥 / 7 + 𝑦 f(x,y)=x/7+y. Answer Partial derivative with respect to 𝑥 x: 𝑓 𝑥 = 1 7 f x = 7 1 Partial derivative with respect to 𝑦 y: 𝑓 𝑦 = 1 f y =1 Question 4 What is the derivative of cot ⁡ 2 ( 𝑥 ) cot 2 (x)? Answer 1 Using the chain rule: 𝑓 ( 𝑥 ) = cot ⁡ 2 ( 𝑥 ) ⟹ 𝑑 𝑑 𝑥 [ cot ⁡ 2 ( 𝑥 ) ] = 2 cot ⁡ ( 𝑥 ) ⋅ ( − csc ⁡ 2 ( 𝑥 ) ) = − 2 cot ⁡ ( 𝑥 ) csc ⁡ 2 ( 𝑥 ) f(x)=cot 2 (x)⟹ dx d [cot 2 (x)]=2cot(x)⋅(−csc 2 (x))=−2cot(x)csc 2 (x) Answer 2 Alternate form: 𝑑 𝑑 𝑥 [ cot ⁡ 2 ( 𝑥 ) ] = 2 cot ⁡ ( 𝑥 ) ⋅ 𝑑 𝑑 𝑥 [ cot ⁡ ( 𝑥 ) ] = 2 cot ⁡ ( 𝑥 ) ( − csc ⁡ 2 ( 𝑥 ) ) = − 2 cot ⁡ ( 𝑥 ) csc ⁡ 2 ( 𝑥 ) dx d [cot 2 (x)]=2cot(x)⋅ dx d [cot(x)]=2cot(x)(−csc 2 (x))=−2cot(x)csc 2 (x) Question 5 What’s the derivative of sec ⁡ 2 ( 𝑥 ) sec 2 (x)? Answer 1 Using the chain rule: 𝑑 𝑑 𝑥 [ sec ⁡ 2 ( 𝑥 ) ] = 2 sec ⁡ ( 𝑥 ) ⋅ sec ⁡ ( 𝑥 ) tan ⁡ ( 𝑥 ) = 2 sec ⁡ 2 ( 𝑥 ) tan ⁡ ( 𝑥 ) dx d [sec 2 (x)]=2sec(x)⋅sec(x)tan(x)=2sec 2 (x)tan(x) Answer 2 Simpler derivation using chain rule: 𝑑 𝑑 𝑥 [ sec ⁡ 2 ( 𝑥 ) ] = 2 sec ⁡ ( 𝑥 ) tan ⁡ ( 𝑥 ) sec ⁡ ( 𝑥 ) = 2 sec ⁡ 2 ( 𝑥 ) tan ⁡ ( 𝑥 ) dx d [sec 2 (x)]=2sec(x)tan(x)sec(x)=2sec 2 (x)tan(x)