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 h(u)=F(x(u))[/tex]<br /> You are asked to find \frac{dh}{du}<br /> <br /> <br /> Note that it is easy to give your final answer in terms of &quot;x&quot; rather than &quot;u&quot;, 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 \frac{df}{du}\frac{du}{dx} in that is the &quot;chain&quot; 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}}