# Differentiation problem

Printable View

• May 21st 2009, 11:49 AM
Ant
Differentiation problem
Hi,

I'm mistified about where I'm going wrong here:

Find the derivative of: [(4x-1)(4x-1)^1/2]

The answer given is 6(4x-1)^1/2

but I get 6(4x -1)

not sure where i'm going wrong if anyone could help that would be great!

Thanks.
• May 21st 2009, 12:06 PM
skeeter
Quote:

Originally Posted by Ant
Hi,

I'm mistified about where I'm going wrong here:

Find the derivative of: [(4x-1)(4x-1)^1/2]

The answer given is 6(4x-1)^1/2

but I get 6(4x -1)

not sure where i'm going wrong if anyone could help that would be great!

Thanks.

$\displaystyle y = (4x-1)(4x-1)^{\frac{1}{2}} = (4x-1)^{\frac{3}{2}}$

$\displaystyle y' = \frac{3}{2}(4x-1)^{\frac{1}{2}} \cdot 4$

$\displaystyle y' = 6(4x-1)^{\frac{1}{2}}$
• May 21st 2009, 12:12 PM
Ant
Thanks, that makes perfect sense.

could you explain what's wrong with my method please?

u = 4x-1
u' = 4
v = (4x-1)^1/2
v' = 2/[(4x-1)^1/2]

using product rule --

4(4X-1)1/2 + 2(4x-1)/[(4x-1)^1/2]

multiplying through by (4x-1)^1/2

4(4x-1) + 2(4x-1) = 24x-6 = 6(4x-1)

I know i'm going wrong somewhere..I just can't see where!
• May 21st 2009, 01:46 PM
skeeter
Quote:

Originally Posted by Ant
Thanks, that makes perfect sense.

could you explain what's wrong with my method please?

u = 4x-1
u' = 4
v = (4x-1)^1/2
v' = 2/[(4x-1)^1/2]

using product rule --

4(4X-1)1/2 + 2(4x-1)/[(4x-1)^1/2]

multiplying through by (4x-1)^1/2

no, you have to multiply through by 1 ...
$\displaystyle \textcolor{red}{\frac{(4x-1)^{\frac{1}{2}}}{(4x-1)^{\frac{1}{2}}}}$

4(4x-1) + 2(4x-1) = 24x-6 = 6(4x-1)

I know i'm going wrong somewhere..I just can't see where!

.
• May 21st 2009, 01:53 PM
Ant
thanks, of course!

I thiink I was trying to treat the plus sign like it was an equals sign...