USING CHAIN RULE AND QUOTIENT RULE TOGETHER

When two differentiable functions are divided, we may have to use quotient rule to find the derivative.

d(u/v) / dx = (vu' - uv')/v2

When one function is defined in terms of another function, we may have to use chain rule.

y = f (g(x))

dy/dx = f'(g(x)) g'(x)

Here we see some examples to find the derivation involving these two methods.

Differentiate (using an embedded chain rule):

Problem 1:

2x/(x + 1)1/2

Solution:

y = 2x/(x + 1)1/2

u = 2x and v = (x + 1)1/2

u' = 2 and v' = 1/2√(x + 1)

y = 2xx+1u = 2x and v=u'=2(1) v'=x+1 => v'=dydx = x+12 - 2x 12x+1x+12dydx = 2x+1 - xx+1x+1dydx = 2(x+1)- x(x+1)x+1dydx = 2x+2- x(x+1)x+1dydx = x+2(x+1)x+1

Problem 2 :

(2x + 7)3/(4x - 1)

Solution:

y = (2x+7)3(4x - 1)u = (2x+7)3 and v=4x - 1u'=3(2x+7)2(2) ⟹ 6(2x+7)2v'=4(1)-0=> v'=4dydx = (4x - 1)6(2x+7)2 - (2x+7)3 (4)(4x - 1)2dydx = 2(2x+7)2[3(4x-1)- 2(2x+7)](4x - 1)2dydx = 2(2x+7)2[12x-3- 4x-14](4x - 1)2dydx = 2(2x+7)2(8x-17)(4x - 1)2

Problem 3 :

(x - 1)/(7x + 3)4

Solution :

Here u = x - 1 and v = (7x + 3)4

u' = 1 and v' = 4(7x + 3)3 (7)

v' = 28(7x + 3)3

Applying into the formula, we get

dydx = (7x+ 3)4(1) - (x - 1) 28 (7x+ 3)3(7x+ 3)42dydx = (7x+ 3)3[7x + 3 - 28(x - 1)](7x+ 3)8dydx = (7x+ 3)3[7x + 3 - 28x + 28)](7x+ 3)8dydx = (31- 14x)(7x+ 3)5

Problem 4 :

(2x - 5)/√(x + 1)

Solution :

Here u = x - 1 and v = (7x + 3)4

u' = 1 and v' = 4(7x + 3)3 (7)

v' = 28(7x + 3)3

Applying into the formula, we get

y = 2x -5x + 1u = 2x - 5 and v = x + 1u' = 2 and v' = 12x + 1 dydx = x + 1 (2) - (2x-5)12x + 1 x + 12dydx = 4(x + 1) - (2x-5)2x + 1 (x + 1)dydx =4 x +4 - 2x+52x + 1(x + 1)dydx =2 x +92(x + 1)x + 1

Problem 5 :

√(x + 1)/(4x + 1)

Solution :

y = x-14x+1u = x-1 and v = u' = 12x-1 and v' = dydx = (4x+1)12x-1 -4x-1 (4x+1)2dydx = (4x + 1) - 8(x-1)2x + 1 (4x+1)2dydx =4x+1-8x+8(4x+1)2x + 1dydx =9-4x(4x+1)2x + 1

Problem 6 :

√(x2 + 1)/(x - 8)2

Solution :

y = x2+1(x-8)2u = x2+1 and v = u' = 12x2+1 (2x) and v' = u' = xx2+1 dydx = (x-8)2xx2+1 -x2+1 2(x-8) (x-8)4dydx = x(x-8)2x2+1 - 2(x-8) x2+1 (x-8)4dydx = x(x-8)2- 2(x-8) x2+1x2+1 (x-8)4dydx =(x-8)x(x-8) - 2x2+1 (x-8)4x2+1dydx =x2-8x -2x2-2(x-8)3x2+1dydx =-x2-8x-2(x-8)3x2+1

Problem 7 :

(x + 4)/3√x

Solution :

y = (x - 4)3xu = x -4 and v = 3xu' =1 and v' = 13x-23dydx =3x (1) - (x - 4) 13x-233x2dydx =x13 - (x-4)3x23x23dydx = 3x13 + 23 - (x - 4)3x23x23dydx = 3x - x + 43x23+23dydx = 2x +43x43

Problem 8 :

(x + 3)4/x2

Solution :

y = (x + 3)4/x2

u = (x + 3)4 and v = x2

u' = 4(x + 3)3 and v' = 2x

dy/dx = [4 x2(x + 3)3 - 2x (x + 3)4]/x4

dy/dx = (x + 3)3[4 x2 - 2x (x + 3)]/x4

dy/dx = (x + 3)3[4 x2- 2x2 - 6x)]/x4

dy/dx = (x + 3)3(2x2 - 6x)/x4

dy/dx = x(x + 3)3(2x - 6)/x4

dy/dx = (x + 3)3(2x - 6)/x3

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