# detailed working for differentiation

• Feb 9th 2009, 12:13 AM
mattty
detailed working for differentiation
Hi, I've been working on my differentiation unit for the last couple of days and I am having a lot of trouble getting the correct answer.

as an example, for the question
$\displaystyle (dy)/(dx)$ of $\displaystyle y= 2/(x^2+1)$
$\displaystyle (-4x)/(x^4+2x^3+x^2+2x+1)$
as apposed to the textbook answer of
$\displaystyle (-4x)/(x^4+2x^2+1)$

I've been looking through my working to find the problem, and i've found the location where i go in a different direction from the proper answer, but i can't see how i would change anything to get the right answer. Since i can't find the right answer with my own working out, my question is could somebody please point out where i go wrong and correct it (preferably in detail.

here is my working out (numerator is unimportant so i am not showing the working for it)

$\displaystyle =(2/((x+h)^2+1))-(2(x^2+1)$
$\displaystyle =((2(x^2+1))/((x^2+1)(x+h)^2+1))-((2((x+h^2)+1))/((x^2+1)(x+h)^2+1))$
$\displaystyle =(-4xh+-2h^2)/((x^2+1)(x+h)^2+1))$
the next 2 lines is where i suspect my mistake to be
$\displaystyle =(-4x+-2h)/(x^2(x^2+2hx+h^2)+1(x^2+2hx+h^2))$
$\displaystyle =(-4x+-2h)/(x^4+2hx^3+x^2h^2+x^2+2hx+h^2+1)$
lim h→0 $\displaystyle (-4x)/(x^4+2x^3+x^2+2x+1)$
$\displaystyle =(-4x)/(x^4+2x^3+x^2+2x+1)$
which does not $\displaystyle =(-4x)/(x^4+2x^2+1)$

Any help that is given would be greatly appreciated.
• Feb 9th 2009, 12:37 AM
Quote:

Originally Posted by mattty
Hi, I've been working on my differentiation unit for the last couple of days and I am having a lot of trouble getting the correct answer.

as an example, for the question
$\displaystyle (dy)/(dx)$ of $\displaystyle y= 2/(x^2+1)$
$\displaystyle (-4x)/(x^4+2x^3+x^2+2x+1)$
as apposed to the textbook answer of
$\displaystyle (-4x)/(x^4+2x^2+1)$

I've been looking through my working to find the problem, and i've found the location where i go in a different direction from the proper answer, but i can't see how i would change anything to get the right answer. Since i can't find the right answer with my own working out, my question is could somebody please point out where i go wrong and correct it (preferably in detail.

here is my working out (numerator is unimportant so i am not showing the working for it)

$\displaystyle =(2/((x+h)^2+1))-(2(x^2+1)$
$\displaystyle =((2(x^2+1))/((x^2+1)(x+h)^2+1))-((2((x+h^2)+1))/((x^2+1)(x+h)^2+1))$
$\displaystyle =(-4xh+-2h^2)/((x^2+1)(x+h)^2+1))$
the next 2 lines is where i suspect my mistake to be
$\displaystyle =(-4x+-2h)/(x^2(x^2+2hx+h^2)+1(x^2+2hx+h^2))$
$\displaystyle =(-4x+-2h)/(x^4+2hx^3+x^2h^2+x^2+2hx+h^2+1)$
lim h→0 $\displaystyle (-4x)/(x^4+2x^3+x^2+2x+1)$
$\displaystyle =(-4x)/(x^4+2x^3+x^2+2x+1)$
which does not $\displaystyle =(-4x)/(x^4+2x^2+1)$

Any help that is given would be greatly appreciated.

Perhaps you can try using the quotient rule .

$\displaystyle y=\frac{2}{x^2+1}$

$\displaystyle u=2 , \frac{du}{dx}=0$

$\displaystyle v=x^2+1 , \frac{dv}{dx}=2x$

Formula : $\displaystyle \frac{v(du/dx)-u(dv/dx)}{v^2}$

Try substituting into the formulae and you should get the answer .
• Feb 9th 2009, 12:46 AM
mattty
Quote:

Perhaps you can try using the quotient rule .

$\displaystyle y=\frac{2}{x^2+1}$

$\displaystyle u=2 , \frac{du}{dx}=0$

$\displaystyle v=x^2+1 , \frac{dv}{dx}=2x$

Formula : $\displaystyle \frac{v(du/dx)-u(dv/dx)}{v^2}$

Try substituting into the formulae and you should get the answer .

Thanks for the forumla, but is there any chance someone could help me do it the long way so I can see where I'm going wrong? Mainly because I don't want to have to use forumlas from the second half of the chapter for questions from the first half, it just doesn't seem right.
• Feb 9th 2009, 01:44 AM
Krizalid
There's no need to complicate! It's $\displaystyle y=\frac{2}{{{x}^{2}}+1}=2{{\left( {{x}^{2}}+1 \right)}^{-1}}.$ Hence $\displaystyle y'=-\frac{4x}{(x^2+1)^2}$ which agrees with your answer.
• Feb 9th 2009, 02:08 AM
mr fantastic
Quote:

Originally Posted by mattty
Hi, I've been working on my differentiation unit for the last couple of days and I am having a lot of trouble getting the correct answer.

as an example, for the question
$\displaystyle (dy)/(dx)$ of $\displaystyle y= 2/(x^2+1)$
$\displaystyle (-4x)/(x^4+2x^3+x^2+2x+1)$
as apposed to the textbook answer of
$\displaystyle (-4x)/(x^4+2x^2+1)$

I've been looking through my working to find the problem, and i've found the location where i go in a different direction from the proper answer, but i can't see how i would change anything to get the right answer. Since i can't find the right answer with my own working out, my question is could somebody please point out where i go wrong and correct it (preferably in detail.

here is my working out (numerator is unimportant so i am not showing the working for it)

$\displaystyle =(2/((x+h)^2+1))-(2(x^2+1)$
$\displaystyle =((2(x^2+1))/((x^2+1)(x+h)^2+1))-((2((x+h^2)+1))/((x^2+1)(x+h)^2+1))$
$\displaystyle =(-4xh+-2h^2)/((x^2+1)(x+h)^2+1))$

Mr F says: $\displaystyle {\color{red} = \frac{-h(4x + 2h)}{[(x + h)^2 + 1] (x^2 + 1)}}$.

Now divide by h: $\displaystyle {\color{red} \frac{-(4x + 2h)}{[(x + h)^2 + 1] (x^2 + 1)}}$.

Now take the limit h --> 0: $\displaystyle {\color{red} \frac{-(4x + 0)}{[(x + 0)^2 + 1] (x^2 + 1)} = \frac{-4x}{(x^2 + 1)^2}}$

and all is well.

What you have done below is to make more work for yourself than is necessary. This wastes time, increases the chance of mistakes etc.

You will no doubt have made the sort of careless mistake that I can never be bothered looking for.

the next 2 lines is where i suspect my mistake to be
$\displaystyle =(-4x+-2h)/(x^2(x^2+2hx+h^2)+1(x^2+2hx+h^2))$
$\displaystyle =(-4x+-2h)/(x^4+2hx^3+x^2h^2+x^2+2hx+h^2+1)$
lim h→0 $\displaystyle (-4x)/(x^4+2x^3+x^2+2x+1)$
$\displaystyle =(-4x)/(x^4+2x^3+x^2+2x+1)$
which does not $\displaystyle =(-4x)/(x^4+2x^2+1)$

Any help that is given would be greatly appreciated.

..
• Feb 9th 2009, 02:41 AM
Krizalid
Ooops, sorry, I see now I misread the question.
• Feb 9th 2009, 04:46 AM
mattty
Quote:

Originally Posted by mr fantastic
..

thanks so much, i can finally see where i was going wong, and actually understand how to do these types of questions properly now. Much appreciated.