# Thread: First three derivatives of arctan

1. ## First three derivatives of arctan

Hi,

I got stuck while trying to calculate the third derivative for arctan.

The second derivative for arctan is $\displaystyle \frac{-2x}{(1+x^2)^2}$
No problem. Then i try to rewrite it as: $\displaystyle -2x*(1+x^2)^{-2}$ and use the product rule. This is what i end up with:

$\displaystyle -2x*(1+x^2)^{-2}$
$\displaystyle -2x*-2(1+x^2)^{-3} *2x - 2*(1+x^2)^{-2}$

which can be rearanged into: $\displaystyle -4x^2 -\frac{2}{(1+x^2)^3} -\frac{2}{(1+x^2)^2}$

and it's wrong. What am i doing wrong?

2. Hello,
Originally Posted by Jones
Hi,

I got stuck while trying to calculate the third derivative for arctan.

The second derivative for arctan is $\displaystyle \frac{-2x}{(1+x^2)^2}$
No problem. Then i try to rewrite it as: $\displaystyle -2x*(1+x^2)^{-2}$ and use the product rule. This is what i end up with:

$\displaystyle -2x*(1+x^2)^{-2}$
$\displaystyle -2x*-2(1+x^2)^{-3} *2x - 2*(1+x^2)^{-2}$

which can be rearanged into: $\displaystyle -4x^2 -\frac{2}{(1+x^2)^3} -\frac{2}{(1+x^2)^2}$

and it's wrong. What am i doing wrong?
What you are doing wrong is not calculations, but brackets !!!!

It's $\displaystyle (-2x)*(-2)*(1+x^2)^{-3}*(2x)-2*(1+x^2)^{-2}$

And this gives $\displaystyle (8x^2)*(1+x^2)^{-3}-2*(1+x^2)^{-2}=\frac{8x^2}{(1+x^2)^3}-\frac{2}{(1+x^2)^2}=\frac{8x^2-2(1+x^2)}{(1+x^2)^3}$

3. Originally Posted by Jones
Hi,

I got stuck while trying to calculate the third derivative for arctan.

The second derivative for arctan is $\displaystyle \frac{-2x}{(1+x^2)^2}$
No problem. Then i try to rewrite it as: $\displaystyle -2x*(1+x^2)^{-2}$ and use the product rule. This is what i end up with:

$\displaystyle -2x*(1+x^2)^{-2}$
$\displaystyle -2x*-2(1+x^2)^{-3} *2x - 2*(1+x^2)^{-2}$

which can be rearanged into: $\displaystyle -4x^2 -\frac{2}{(1+x^2)^3} -\frac{2}{(1+x^2)^2}$

and it's wrong. What am i doing wrong?
No, it can't be rearranged into that! It is
$\displaystyle \frac{2x^2}{(1+x^2)^3}- \frac{2}{(1+x^2)^2}$
which can be further written as $\displaystyle \frac{2x^2}{(1+x^2)^3}- \frac{2(1+ x^2)}{(1+x^2)^3}= -\frac{2}{(1+x^2)^3}$
You seem to have misread $\displaystyle -2x*-2(1+x^2)^{-3}*2x$ as
$\displaystyle (-2x- 2(1+x^2)^{-3})*2x$, missing the first "*". It is, in fact, [/tex](-2x)(-2)(1+x^2)^{-3}(2x)[/tex]

4. Ok, thanks guys. But are you sure thats correct?

The key says $\displaystyle \frac{-2+6x^2}{(1+x^2)^3}$

look:
Arctan @ pici.se

5. Hi

We are ok with the second derivative $\displaystyle f'' = \frac{-2x}{(1+x^{2})^{2}}$

Now, for the third derivative: Using quotient rule:

$\displaystyle \frac{(1+x^{2})^{2}\cdot (-2) + 2x\cdot (1+x^{2})\cdot 2x}{(1+x^{2})^{4}}$ Simplify by dividing out one $\displaystyle (1+x^{2})$ term.

$\displaystyle \frac{-2\cdot (1+x^{2}) + 8x^{2}}{(1+x^{2})^{3}} = \frac{-2+6x^{2}}{(1+x^{2})^{3}}$

6. Of course, stupid me.
It's the same as Moo showed, it just hadn't been simplified.

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