- #1
bkvitha
- 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.
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
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
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