PRACTICE PROBLEMS ON DIFFERENTIATION USING CHAIN RULE

Problem 1 :

y = (x³ + 3)⁵

Solution :

y = (x³ + 3)⁵

dy/dx = 5(x³ + 3)⁴ (3x² + 0)

= 15 x²(x³ + 3)⁴ 

Problem 2 :

y = (-3x⁵ + 1)³

Solution :

y = (-3x⁵ + 1)³

dy/dx = 3(-3x⁵ + 1)²  (-15x⁴ + 0)

= -45x⁴ (-3x⁵ + 1)²  

Problem 3 :

y = (-5x³ - 3)³

Solution :

y = (-3x⁵ + 1)³

dy/dx = 3(-3x⁵ + 1)²  (-15x⁴ + 0)

= -45x⁴ (-3x⁵ + 1)²  

Problem 4 :

y = ∜(-3x⁴ - 2)

Solution :

y = ∜(-3x⁴ - 2)

Problem 5 :

Solution :

Problem 6 :

Solution :

Problem 7 :

Solution :

Problem 8 :

y = (3x³ + 1) (-4x² - 3)⁴

Solution :

Since two terms are multiplying, we have to use the product rule to find the differentiation.

Let u = 3x³ + 1 and v = (-4x² - 3)⁴

u' = 9x² + 0 and v' = 4(-4x² - 3)³(-8x - 0)

u' = 9x² and v' = -32x(-4x² - 3)³

dy/dx = uv' + vu'

= (3x³ + 1) (-32x(-4x² - 3)³) + (-4x² - 3)⁴ (9x²)

= -32x(3x³ + 1) (-4x² - 3)³ + (9x²) (-4x² - 3)⁴ 

= x(-4x² - 3)³ [-32(3x³ + 1) + (9x) (-4x² - 3)]

= x(-4x² - 3)³ [-96x³ - 32 - 36x³ - 27x]

= x(-4x² - 3)³ [-132x³ - 27x - 32]

Problem 9 :

y = (x³ + 4)⁵ / (3x⁴ - 2)

Solution :

Since two functions are divided, we have to use the quotient rule to find the differentiation.

Let u = (x³ + 4)⁵ and v = (3x⁴ - 2)

u' = 5(x³ + 4)⁴(3x² + 0)

u' = 15x²(x³ + 4)⁴

v' = (12x³ - 0)

dy/dx = (vu' - uv')/v²

Problem 10 :

y = ((x + 5)⁵ - 1)⁴ 

Solution :

y = ((x + 5)⁵ - 1)⁴ 

dy/dx = 4((x + 5)⁵ - 1)³ (5(x + 5)⁴ - 0) (1)

= 20[ (x + 5)⁵ - 1)³ (x + 5)⁴ ]

= 20(x + 5)⁴ [ ((x + 5)⁵ - 1)³ ]