# Finding the gradient function.... lost

• July 29th 2009, 02:55 AM
Joel
Finding the gradient function.... lost
Hi all,

Firstly I am asked to differentiate y= x^4 - 3x^2 + 6

I get 4x^3 - 6x I am sure this correct, please advise if not (Nod)

I am lost with the gradient function of this one

y = 3/x^2 - 2x + 5

not sure what to do with the initial fraction

thinking it should be

????? - 2 ??

Cheers
• July 29th 2009, 03:14 AM
Plato
Quote:

Originally Posted by Joel
the gradient function of this one
y = 3/x^2 - 2x + 5

If $y=\frac{c}{f(x)}$ then $y'=\frac{-cf'(x)}{(f(x))^2}$
• July 29th 2009, 03:20 AM
Joel
If http://www.mathhelpforum.com/math-he...93532f8d-1.gif then http://www.mathhelpforum.com/math-he...02eda108-1.gif

Therefore

y = 3/x^2 - 2x + 5

dy/dx -3 (2x) / (x^2)^2 -2

-6x/x^4 -2 ????

Is this correct?
• July 29th 2009, 04:39 AM
mr fantastic
Quote:

Originally Posted by Joel
Hi all,

Firstly I am asked to differentiate y= x^4 - 3x^2 + 6

I get 4x^3 - 6x I am sure this correct, please advise if not (Nod)

I am lost with the gradient function of this one

y = 3/x^2 - 2x + 5

not sure what to do with the initial fraction

thinking it should be

????? - 2 ??

Cheers

You need to be less ambiguous. Is it $y = \frac{3}{x^2} - 2x + 5$ or $y = \frac{3}{x^2 - 2x + 5}$?
• July 29th 2009, 04:43 AM
Joel
• July 29th 2009, 05:58 AM
HallsofIvy
Quote:

Originally Posted by Joel

Then $y= 3x^{-2}- 2x+ 5$. That's easier than using the quotient rule.
• July 29th 2009, 06:01 AM
Joel
http://www.mathhelpforum.com/math-he...ec4f5e34-1.gif.

this would make

y'' = -6xˉ³ - 2 ??