Differentiate !?

Question

y = (2x+3)^3 / root4x-7

the root4x-7 = (4x-7)^1/2 i just didnt know how to write it haha

i know you use quotiant rule and chain rule. I get up to this:
= [3(2x+3)^2 (4x-7)^1/2 - 1/2(4x-7)^-1/2 (2x+3)^3 / 4x-7 ] 8
am i on the right track ?
haha thanks =)

Answer 1

y = (2x + 3)^3 / √(4x - 7)
ln (y) = 3*ln (2x + 3) - 0.5*ln (4x - 7)
(1/y)(dy/dx) = 3*2 / (2x + 3) - 0.5*4 / (4x - 7)
dy/dx = [6/(2x + 3) - 2/(4x - 7)]y
dy/dx = [6/(2x + 3) - 2/(4x - 7)][(2x + 3)^3 / √(4x - 7)]
dy/dx = [{6(2x + 3)^2 / √(4x - 7)} - {2(2x + 3)^3}√(4x - 7)]

In such type of problems it is wiser to take natural logarithm of both sides and then differentiate. This reduces the algebraic calculations.

Answer 2

y = (2x + 3)³ / (4x - 7)^(1/2) = u / v
u = (2x + 3)³
v = (4x - 7)^(1/2)
du/dx = 3 (2x + 3)² (2) = 6 (2x + 3)²
dv/dx = (1/2)(4x - 7)^(-1/2)(4)
dv/dx = 2 (4x - 7)^(-1/2)
dy / dx = (v du /dx - u dv / dx) / v ²
dy/dx = N / D

N is given by:-
v du/dx - u dv/dx
(4x - 7)^(1/2) (6) (2x + 3)²
- (2x + 3)³ (2) (4x - 7)^(-1/2)
6(4x - 7)^(1/2)(2x + 3)² - 2(4x - 7)^(-1/2)(2x + 3)³
2(2x + 3)² (4x - 7)^(-1/2) [ 3(4x - 7) - (2x + 3) ]
2 (2x + 3)² (4x - 7)^(-1/2) (10x - 24)

D is given by:-
(4x - 7)

N / D is given by:-
2 (2x + 3)²(10x - 24) / (4x - 7)^(3/2)
4 (2x + 3)² (5x - 12) / (4x - 7)^(3/2)

Answer 3

lny = 3ln(2x+3) - (1/2)ln(4x-7)
y'/y = 6/(2x+3) - 2/(4x-7)
y' = [(2x+3)^3 /sqrt(4x-7)][ 6/(2x+3) - 2/(4x-7)]
= [(2x+3)^2 /(4x-7)^(3/2)][4(5x-12)]
= 4(2x+3)^2 (5x-12)/(4x-7)^(3/2)
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To the answerer below me,
(8x - 27) is not a factor. Double check it, please!