Differentiate y = 4x^3 * sqrt.(2x-1)

[Monkeyseat]

Hi,

Question: Differentiate y = 4x^3 * sqrt.(2x-1).

That wasn't a problem, I differentiated that and got dy/dx = ((4x^3)/(sqrt.(2x-1))) + 12x^2 * sqrt.(2x-1).

I know that's correct. The question asks you to show it in this form however:

(4x^2(7x-3))/(sqrt.(2x-1))

I've been looking at this for ages and I just can't see how I can change ((4x^3)/(sqrt.(2x-1))) + 12x^2 * sqrt.(2x-1) into (4x^2(7x-3))/(sqrt.(2x-1))...

Any tips?

Thanks.

Btw sorry for the bad formatting.

 

[Pklarreich]

Hi,

Question: Differentiate y = 4x^3 * sqrt.(2x-1).

That wasn't a problem, I differentiated that and got dy/dx = ((4x^3)/(sqrt.(2x-1))) + 12x^2 * sqrt.(2x-1).

I know that's correct. The question asks you to show it in this form however:

(4x^2(7x-3))/(sqrt.(2x-1))

I've been looking at this for ages and I just can't see how I can change ((4x^3)/(sqrt.(2x-1))) + 12x^2 * sqrt.(2x-1) into (4x^2(7x-3))/(sqrt.(2x-1))...

Any tips?

Thanks.

Btw sorry for the bad formatting.

You have:

4x^3
---------- + 12x^2 sqrt.(2x-1)
sqrt.(2x-1)

Looks like a standard 'combine fractions over an LCD' example.

4x^3 + 12x^2 (2x-1)
------------------------
sqrt.(2x-1)

Common factor of 4x^2 on top:

4x^2(x + 3(2x-1))
-------------------
sqrt.(2x-1)

I think that does it.

 

[Monkeyseat]

Thanks for helping but that's not the right answer. The book says:

(4x^2(7x-3))/(sqrt.(2x-1))

The book could be wrong though.

I might have forgot some basic arithmatic but how come you can put 12x^2 sqrt.(2x-1) over sqrt.(2x-1)?

 

[Monkeyseat]

Thanks for helping but that's not the right answer. The book says:

(4x^2(7x-3))/(sqrt.(2x-1))

The book could be wrong though.

I might have forgot some basic arithmatic but how come you can put 12x^2 sqrt.(2x-1) over sqrt.(2x-1)?

Ooops, sorry, you are correct. In fact, both are, they are just wrote in a different form. I get what you've done but how can I write what you said as:

(4x^2(7x-3))/(sqrt.(2x-1))

EDIT:

Nevermind, I got there eventually! Thanks.