Arctan Derivatives Problem: How to Differentiate with Respect to Arctan x?

[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??

 

[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 [itex]h(u)=F(x(u))[/tex]
You are asked to find $\frac{dh}{du}$


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

 

[HallsofIvy]

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 [itex]\frac{df}{du}\frac{du}{dx} in that is the "chain" rule.

 

[teng125]

sotty,for threat 2 i don't really understand.can u plsexplain more clearly??

 

[arildno]

HallsofIvy's "u" is not the same as my "u".
He is presenting a technique to evaluate the term $\frac{dF}{dx}$ 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 $\frac{dh}{du}$

 

[teng125]

can u pls show me the whole steps ??thanx

 

[arildno]

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

 

[Calin]

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

 

[VietDao29]

can u pls show me the whole steps ??thanx

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.

 

[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:
$$\frac{dF}{df}=\frac{\frac{dF}{dx}}{\frac{df}{dx}}$$