Arctan derivatives problem

1. Feb 11, 2006

teng125

Find the derivatives of the following arctan {[(1+x^2)^1/2 ] - 1} / x with respect to arctan x.may i know how to do this??
how to start and the steps??

2. Feb 11, 2006

arildno

Let your independent variable be given as $u=arctan(x)$ that is,
$x(u)=tan(u)$

You have been given the function of x,
$$F(x)=\frac{arctan(\sqrt{1+x^{2}}-1)}{x}$$
Let $h(u)=F(x(u))[/tex] You are asked to find [itex]\frac{dh}{du}$

Note that it is easy to give your final answer in terms of "x" rather than "u", since we have $$\frac{dx}{du}=\frac{1}{\cos^{2}u}=1+\tan^{2}u=1+x^{2}$$

Last edited: Feb 11, 2006
3. Feb 11, 2006

HallsofIvy

Staff Emeritus
You have $\frac{arctan(u)}{x}$. Since that is a quotient of two functions use the "quotient" rule:
$$\frac{d\left(\frac{f(u)}{x}}= \frac{\frac{df}{du}\frac{du}{dx}x- f(u)}{x}$$
where $u= \sqrt{1+x^2}-1$ and f(u)= arctan u.
The $\frac{df}{du}\frac{du}{dx} in that is the "chain" rule. 4. Feb 11, 2006 teng125 sotty,for threat 2 i don't really understand.can u plsexplain more clearly?? 5. Feb 11, 2006 arildno HallsofIvy's "u" is not the same as my "u". He is presenting a technique to evaluate the term [itex]\frac{dF}{dx}$ in my terms.

We have: $\frac{d}{dx}(\sqrt{1+x^{2}}-1)=\frac{x}{\sqrt{1+x^{2}}}$
Thus, we get:
$$\frac{dF}{dx}=\frac{\sqrt{1+x^{2}}}{1+(\sqrt{1+x^{2}}-1)^{2}}-\frac{F(x)}{x}$$
Multiply this with $1+x^{2}$ to get $\frac{dh}{du}$

6. Feb 11, 2006

teng125

can u pls show me the whole steps ??thanx

7. Feb 11, 2006

arildno

Darn, I made a mistake!
the correct expression for the derivative with respect to u=arctan(x), should be:
$$\frac{dh}{du}=\frac{\sqrt{1+x^{2}}}{1+(\sqrt{1+x^{2}}-1)^{2}}-F(x)\frac{1+x^{2}}{x}$$

8. Feb 11, 2006

Calin

$\\int_1^4 \\sqrt{t}(ln(t))dt$

9. Feb 11, 2006

VietDao29

Unfortunately, I don't think we are going to provide you a step by step solution. However, we may help you.
Just tell us where in the post that you don't really understand and we may clarify it for you.

10. Feb 12, 2006

arildno

It might be helpful to remember the following rule:

Suppose you are to differentiate a function F(x) with respect to a (invertible) function f(x).
Then we have, in general:
$$\frac{dF}{df}=\frac{\frac{dF}{dx}}{\frac{df}{dx}}$$