Find the derivative of the question

[ASidd]

Homework Statement

(9x-8)/ 2(sqrt(3x-4))

Homework Equations

Vu'-Uv'/v^2

i.e. quotient rule

The Attempt at a Solution

I get
V=2(sqrt(3x-4) u=9x-8
V'=3((3x-4)^-1.5) u'=9

When I put this into the equation and solve I get
9x-8(sqrt(3x-4))/6x-8

But the answer is 3(9x-16)/4((3x-4)^1.5)

Help Please?

 

[HallsofIvy]

Homework Statement

(9x-8)/ 2(sqrt(3x-4))

Homework Equations

Vu'-Uv'/v^2

i.e. quotient rule

The Attempt at a Solution

I get
V=2(sqrt(3x-4) u=9x-8
V'=3((3x-4)^-1.5) u'=9

When I put this into the equation and solve I get
9x-8(sqrt(3x-4))/6x-8

With v= 2sqrt(3x- 4)= 2(3x- 4)^1/2, v'= 3(3x- 4)^-(1/2)= 3(3x- 4)^(.5-1)= 3(3x- 4)^-.5, not "-1.5". That's your basic error.

So u'v- uv'= 9(2(3x-4)^1/2)- (9x- 8)(3(3x-4)^-1/2
To combine those two square roots, use the fact that (3x-4)^1/2= (3x- 4)(3x-4)^-1/2 so we can factor (3x- 4)^-1/2 out: (3x-4)^-1/2(18(3x- 4)- 3(9x- 8))= (3x-4)^-1/2(54x- 72- 27x+ 24)= (3x- 4)^-1/2(27x- 48)= (3x- 4)^-1/2(3)(9x- 16).

Can you finish?

But the answer is 3(9x-16)/4((3x-4)^1.5)

Help Please?

By the way, with that power in the denominator, so you will need to use the chain rule, anyway, it might be simpler to write the function as (1/2)(9x- 8)(3x-4)^-1/2 and use the product rule rather than the quotient rule.

 

[HomogenousCow]

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