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

2024
Views:

55

Pages:

2
Author:

Javon Neal

Question 1 Find the derivatives of the function 𝑔 ( 𝑥 ) = tan ⁡ 3 𝑥 ( 𝑥 + 7 ) 4 g(x)= (x+7) 4 tan3x . Answer Use the quotient rule: 𝑑 𝑑 𝑥 ( 𝑢 𝑣 ) = 𝑣 𝑑 𝑢 𝑑 𝑥 − 𝑢 𝑑 𝑣 𝑑 𝑥 𝑣 2 . dx d ( v u )= v 2 v dx du −u dx dv . Let 𝑢 = tan ⁡ 3 𝑥 u=tan3x and 𝑣 = ( 𝑥 + 7 ) 4 v=(x+7) 4 . Differentiate 𝑢 u and 𝑣 v: 𝑢 = tan ⁡ 3 𝑥 ⟹ 𝑑 𝑢 𝑑 𝑥 = 3 sec ⁡ 2 ( 3 𝑥 ) u=tan3x⟹ dx du =3sec 2 (3x), 𝑣 = ( 𝑥 + 7 ) 4 ⟹ 𝑑 𝑣 𝑑 𝑥 = 4 ( 𝑥 + 7 ) 3 v=(x+7) 4 ⟹ dx dv =4(x+7) 3 . Substitute into the quotient rule: 𝑔 ′ ( 𝑥 ) = ( 𝑥 + 7 ) 4 ⋅ 3 sec ⁡ 2 ( 3 𝑥 ) − tan ⁡ ( 3 𝑥 ) ⋅ 4 ( 𝑥 + 7 ) 3 ( 𝑥 + 7 ) 8 . g ′ (x)= (x+7) 8 (x+7) 4 ⋅3sec 2 (3x)−tan(3x)⋅4(x+7) 3 . Simplify: 𝑔 ′ ( 𝑥 ) = 3 ( 𝑥 + 7 ) 3 sec ⁡ 2 ( 3 𝑥 ) − 4 tan ⁡ ( 3 𝑥 ) ( 𝑥 + 7 ) 5 . g ′ (x)= (x+7) 5 3(x+7) 3 sec 2 (3x)−4tan(3x) . Question 2 What is the instantaneous velocity of an object moving in accordance to 𝑓 ( 𝑡 ) = ( 𝑡 2 sin ⁡ ( 𝑡 − 𝜋 ) , 𝑡 cos ⁡ 𝑡 ) f(t)=(t 2 sin(t−π),tcost) at 𝑡 = 𝜋 3 t= 3 π ? Answer For 𝑥 ( 𝑡 ) = 𝑡 2 sin ⁡ ( 𝑡 − 𝜋 ) x(t)=t 2 sin(t−π), use the product rule: 𝑥 ′ ( 𝑡 ) = 2 𝑡 sin ⁡ ( 𝑡 − 𝜋 ) + 𝑡 2 cos ⁡ ( 𝑡 − 𝜋 ) . x ′ (t)=2tsin(t−π)+t 2 cos(t−π). For 𝑦 ( 𝑡 ) = 𝑡 cos ⁡ 𝑡 y(t)=tcost, differentiate using the product rule: 𝑦 ′ ( 𝑡 ) = cos ⁡ 𝑡 − 𝑡 sin ⁡ 𝑡 . y ′ (t)=cost−tsint. Find the derivative of the parametric curve 𝑓 ′ ( 𝑡 ) f ′ (t): 𝑓 ′ ( 𝑡 ) = 𝑦 ′ ( 𝑡 ) 𝑥 ′ ( 𝑡 ) . f ′ (t)= x ′ (t) y ′ (t) . At 𝑡 = 𝜋 3 t= 3 π , substitute: sin ⁡ ( 𝜋 3 − 𝜋 ) = − sin ⁡ ( 𝜋 3 ) sin( 3 π −π)=−sin( 3 π ), cos ⁡ ( 𝜋 3 ) = 1 2 cos( 3 π )= 2 1 , sin ⁡ ( 𝜋 3 ) = 3 2 sin( 3 π )= 2 3 . Simplify: 𝑓 ′ ( 𝜋 3 ) = cos ⁡ ( 𝜋 3 ) − 𝜋 3 sin ⁡ ( 𝜋 3 ) 2 ( 𝜋 3 ) sin ⁡ ( 𝜋 3 ) + ( 𝜋 3 ) 2 cos ⁡ ( 𝜋 3 ) . f ′ ( 3 π )= 2( 3 π )sin( 3 π )+( 3 π ) 2 cos( 3 π ) cos( 3 π )− 3 π sin( 3 π ) . Evaluate numerically: 𝑓 ′ ( 𝜋 3 ) ≈ 0.172261. f ′ ( 3 π )≈0.172261. Question 3 Find the derivative of the function 𝑓 ( 𝑥 ) = 7 + 𝑥 sec ⁡ 𝑥 f(x)= 7+xsecx . Answer Write the function in power form: 𝑓 ( 𝑥 ) = ( 7 + 𝑥 sec ⁡ 𝑥 ) 1 / 2 . f(x)=(7+xsecx) 1/2 . Differentiate using the chain rule: 𝑓 ′ ( 𝑥 ) = 1 2 ( 7 + 𝑥 sec ⁡ 𝑥 ) − 1 / 2 ⋅ 𝑑 𝑑 𝑥 ( 7 + 𝑥 sec ⁡ 𝑥 ) . f ′ (x)= 2 1 (7+xsecx) −1/2 ⋅ dx d (7+xsecx). Differentiate the inner function 7 + 𝑥 sec ⁡ 𝑥 7+xsecx: 𝑑 𝑑 𝑥 ( 7 + 𝑥 sec ⁡ 𝑥 ) = sec ⁡ 𝑥 ⋅ 𝑑 𝑑 𝑥 ( 𝑥 ) + 𝑥 ⋅ 𝑑 𝑑 𝑥 ( sec ⁡ 𝑥 ) . dx d (7+xsecx)=secx⋅ dx d (x)+x⋅ dx d (secx). Using 𝑑 𝑑 𝑥 ( sec ⁡ 𝑥 ) = sec ⁡ 𝑥 tan ⁡ 𝑥 dx d (secx)=secxtanx: 𝑑 𝑑 𝑥 ( 7 + 𝑥 sec ⁡ 𝑥 ) = sec ⁡ 𝑥 + 𝑥 sec ⁡ 𝑥 tan ⁡ 𝑥 . dx d (7+xsecx)=secx+xsecxtanx. Combine results: 𝑓 ′ ( 𝑥 ) = 1 2 ( 7 + 𝑥 sec ⁡ 𝑥 ) − 1 / 2 ⋅ ( sec ⁡ 𝑥 + 𝑥 sec ⁡ 𝑥 tan ⁡ 𝑥 ) . f ′ (x)= 2 1 (7+xsecx) −1/2 ⋅(secx+xsecxtanx). Simplify the expression: 𝑓 ′ ( 𝑥 ) = sec ⁡ 𝑥 + 𝑥 sec ⁡ 𝑥 tan ⁡ 𝑥 2 7 + 𝑥 sec ⁡ 𝑥 . f ′ (x)= 2 7+xsecx secx+xsecxtanx .