- #1

- 77

- 0

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.

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

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: