HOW TO FIND THE DERIVATIVE OF A SQUARE ROOT FUNCTION

Calculate dy/dx for the following functions:

Problem 1 :

y = (x - 3) √(x - 3)

Solution :

y = (x - 3) √(x - 3)

y = (x - 3) (x - 3)1/2

y = (x - 3)3/2

dy/dx = 3/2 (x - 3)(3/2-1) d/dx (x - 3)

= 3/2 (x - 3)1/2 (1 - 0)

dy/dx = 3√(x - 3)/2

Problem 2 :

y = √(4 - √(x + 2))

Solution :

y = √(4 - √(x + 2))

Problem 3 :

y = (7x + √x² + 3)6

Solution : 

 y = (7x + √x² + 3)6

Problem 4 :

f(x) = √(2x +5)/(7x - 9)

Solution : 

Problem 5 :

f(x) = √(5x3 - 2x)

Solution : 

f(x) = √(5x3 - 2x)

Problem 6 :

y = 3x√(4x2 - 5x + 1)

Solution : 

y = 3x√(4x2 - 5x + 1)

y =3x2 4x2-5x + 1u = 3x2, u' = 6xv = 4x2-5x + 1, v' = 12 4x2-5x + 1 (8x - 5)v' = (8x - 5)2 4x2-5x + 1 dydx =3x2 (8x - 5)2 4x2-5x + 1 +6x 4x2-5x + 1

Problem 7 :

y = (2/3)t√(3t3 - 5t)

Solution : 

y = (2/3)t√(3t3 - 5t)

u = (2/3)t3

u' = (2/3) (3t2)

v = √(3t3 - 5t)

v' = [1/2√(3t3 - 5t)] (9t2 - 5)

v' = [(9t2 - 5)/2√(3t3 - 5t)] 

(2/3)t3[(9t2 - 5)/2√(3t3 - 5t)] + √(3t3 - 5t) (2/3) (3t2)

[t3(9t2 - 5)/3√(3t3 - 5t)] + 2t2√(3t3 - 5t)

Problem 8 :

y = 2ex √x

Solution : 

y = 2ex √x

u = ex    v = √x

u' = ex    v' = 1/2√x

dy/dx = (ex)(1/2√x) + √xex

Factoring ex

dy/dx = (ex)[(1/2√x) + √x]

Problem 9 :

e√(2x^3 - x^2)

Solution :

y =

Problem 10 :

8x5 - 7x4 + 5/x

Solution :

y =

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