Factorising a differntiated equation

1. Sep 26, 2007

bkvitha

Hello there,

I'm quite weak in factorising functions,especially those with indices.
I would appreciate any link on these kind of factorisation or even some tutoring on it.

this is a question I'm stuck in.

y=(2x+3)(4x-3)^1/2, show dy/dx can be written in the form kx/(4x-3)^1/2

my solution:

dy/dx =v(du/dx) + u(dv/dx)

where u=2x+3
du/dx=2
v=(4x-3)^1/2
dv/dx= -2(4x-3)^-1/2

then, dy/dx=[(4x-3)^1/2 ]2 -(-2)(2x+3)[(4x-3)^-1/2]
= 2{(4x-3)^1/2 - (2x+3)(4x-3)^-1/2}

= . . . .

(stumped!)
I am not sure what to do after that

Last edited: Sep 26, 2007
2. Sep 26, 2007

HallsofIvy

Staff Emeritus
There is no question there at all! Do you mean you are asked to find dy/dx in that form?

No, dv/dx= (1/2)(4x-3)^(-1/2)

First, as I said, dv/dx= (1/2)(4x-3)^(-1/2), not what you have.

Making that correction, dy/dx= 2(4x-3)^(1/2)+ (1/2)(2x+3)(4x-3)^(-1/2).

It might help you to note that (4x-3)^(1/2)= (4x-3)/(4x-3)^(1/2) and, of course, (4x-3)^(-1/2)= 1/(4x-3)^(1/2). In other words, by replacing (4x-3)^(1/2) with 2(4x-3)/2(4x-3)^(1/2) you have "common denominators":

dy/dx= 4(4x-3)/(2(4x-3)^(1/2))+ (2x+3)/(2(4x-3)^(-1/2)
can you finish that?

3. Sep 26, 2007

bkvitha

I do not understand why is it so

does not the chain rule state that dy/dx=dy/du *du/dx

hence, when y= U^n

dy/dx= (nu^n-1) * (du/dx)

so

v = (4x-3)^1/2

then dv/dx=1/2(4x-3)^(1/2 -1 ) *(4)
=2(4x-3)^-1/2

(correction to my previous equation with that was dv/dx =-2(4x-3)^-1/2)

correct me if I am wrong, pls n ty!

and the second part.....
I have never learnt that till now that is my Secondory 5 year!(o-levels like)

enlighten me!

4. Sep 26, 2007

bkvitha

Oh oh,
i got the last part , now(had to work it out in a piece of paper first...oops)

5. Sep 26, 2007

bkvitha

but i still do not get the chain rule.
my text book says that , by the waY.