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

2024
Views:

315

Pages:

4
Author:

Javon Neal

Question 1 How do you find the derivative of 𝑦 = ln ⁡ ( 5 𝑥 ) y=ln(5x)? Answer 1 Using the chain rule: 𝑦 = ln ⁡ ( 5 𝑥 ) ⟹ 𝑦 ′ = 1 5 𝑥 ⋅ 5 = 1 𝑥 . y=ln(5x)⟹y ′ = 5x 1 ⋅5= x 1 . Answer 2 By properties of logarithms: 𝑦 = ln ⁡ ( 5 𝑥 ) = ln ⁡ ( 5 ) + ln ⁡ ( 𝑥 ) . y=ln(5x)=ln(5)+ln(x). Since the derivative of a constant ln ⁡ ( 5 ) ln(5) is 0, 𝑑 𝑦 𝑑 𝑥 = 𝑑 𝑑 𝑥 ( ln ⁡ ( 𝑥 ) ) = 1 𝑥 . dx dy = dx d (ln(x))= x 1 . Answer 3 Directly applying the chain rule: 𝑦 ′ = 1 5 𝑥 ⋅ 5 = 1 𝑥 . y ′ = 5x 1 ⋅5= x 1 . Question 2 Find the derivative of 𝑦 = cos ⁡ 2 ( 𝑥 ) y=cos 2 (x). Answer 1 Using the chain rule: 𝑦 = cos ⁡ 2 ( 𝑥 ) = ( cos ⁡ ( 𝑥 ) ) 2 . y=cos 2 (x)=(cos(x)) 2 . Differentiate: 𝑦 ′ = 2 cos ⁡ ( 𝑥 ) ⋅ ( − sin ⁡ ( 𝑥 ) ) = − 2 cos ⁡ ( 𝑥 ) sin ⁡ ( 𝑥 ) . y ′ =2cos(x)⋅(−sin(x))=−2cos(x)sin(x). Simplify: 𝑦 ′ = − sin ⁡ ( 2 𝑥 ) . y ′ =−sin(2x). Answer 2 Alternatively, using trigonometric identities: 𝑦 = cos ⁡ 2 ( 𝑥 ) = 1 2 ( 1 + cos ⁡ ( 2 𝑥 ) ) . y=cos 2 (x)= 2 1 (1+cos(2x)). Differentiate: 𝑦 ′ = 1 2 ⋅ ( − sin ⁡ ( 2 𝑥 ) ) ⋅ 2 = − sin ⁡ ( 2 𝑥 ) . y ′ = 2 1 ⋅(−sin(2x))⋅2=−sin(2x). Answer 3 Apply the chain rule directly: 𝑑 𝑑 𝑥 ( cos ⁡ 2 ( 𝑥 ) ) = 2 cos ⁡ ( 𝑥 ) ( − sin ⁡ ( 𝑥 ) ) = − sin ⁡ ( 2 𝑥 ) . dx d (cos 2 (x))=2cos(x)(−sin(x))=−sin(2x). Question 3 What is the Mixed Derivative Theorem for mixed second-order partial derivatives? How can it help in calculating partial derivatives? Answer 1 The Mixed Derivative Theorem states: If 𝑓 ( 𝑥 , 𝑦 ) f(x,y) and its partial derivatives 𝑓 𝑥 f x , 𝑓 𝑦 f y , 𝑓 𝑥 𝑦 f xy , and 𝑓 𝑦 𝑥 f yx are continuous in an open region, then: 𝑓 𝑥 𝑦 = 𝑓 𝑦 𝑥 . f xy =f yx . This allows us to calculate mixed partial derivatives efficiently by interchanging the order of differentiation. Example: Given 𝑓 ( 𝑥 , 𝑦 ) = 𝑥 3 𝑦 2 + 2 𝑥 𝑦 f(x,y)=x 3 y 2 +2xy: First-order derivatives: ∂ 𝑓 ∂ 𝑥 = 3 𝑥 2 𝑦 2 + 2 𝑦 , ∂ 𝑓 ∂ 𝑦 = 2 𝑥 3 𝑦 + 2 𝑥 . ∂x ∂f =3x 2 y 2 +2y, ∂y ∂f =2x 3 y+2x. Mixed partials: ∂ 2 𝑓 ∂ 𝑥 ∂ 𝑦 = 6 𝑥 𝑦 2 + 2 , ∂ 2 𝑓 ∂ 𝑦 ∂ 𝑥 = 6 𝑥 𝑦 2 + 2. ∂x∂y ∂ 2 f =6xy 2 +2, ∂y∂x ∂ 2 f =6xy 2 +2. Thus, 𝑓 𝑥 𝑦 = 𝑓 𝑦 𝑥 f xy =f yx . Answer 2 Mathematically: ∂ 2 𝑓 ∂ 𝑥 ∂ 𝑦 = ∂ 2 𝑓 ∂ 𝑦 ∂ 𝑥 . ∂x∂y ∂ 2 f = ∂y∂x ∂ 2 f . This simplifies the calculation of higher-order derivatives by reducing redundant steps. Question 4 How are derivatives and one-sided derivatives related? Answer Step 1 One-sided derivatives use a one-sided limit instead of a two-sided limit. Step 2 For a function 𝑓 f, one-sided derivatives at 𝑥 = 𝑎 x=a are: Left derivative: lim ⁡ 𝑥 → 𝑎 − 𝑓 ( 𝑥 ) − 𝑓 ( 𝑎 ) 𝑥 − 𝑎 . x→a − lim x−a f(x)−f(a) . Right derivative: lim ⁡ 𝑥 → 𝑎 + 𝑓 ( 𝑥 ) − 𝑓 ( 𝑎 ) 𝑥 − 𝑎 . x→a + lim x−a f(x)−f(a) . Step 3 If both one-sided derivatives are equal, then the function has a derivative at 𝑥 = 𝑎 x=a: 𝑓 ′ ( 𝑥 ) = lim ⁡ 𝑥 → 𝑎 𝑓 ( 𝑥 ) − 𝑓 ( 𝑎 ) 𝑥 − 𝑎 . f ′ (x)= x→a lim x−a f(x)−f(a) . Question 5 If a function is decreasing with upward concavity, what do you know about the function’s first two derivatives? Answer If a function is decreasing with upward concavity: The first derivative 𝑓 ′ ( 𝑥 ) f ′ (x) is negative because the function is decreasing. The second derivative 𝑓 ′ ′ ( 𝑥 ) f ′′ (x) is positive because the function has upward concavity (convex shape).