derivative

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 [itex]u=arctan(x)[/itex] that is,
[itex]x(u)=tan(u)[/itex]

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


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

HallsofIvy

You have [itex]\frac{arctan(u)}{x}[/itex]. Since that is a quotient of two functions use the "quotient" rule:
[tex]\frac{d\left(\frac{f(u)}{x}}= \frac{\frac{df}{du}\frac{du}{dx}x- f(u)}{x}[/tex]
where [itex]u= \sqrt{1+x^2}-1[/itex] 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 [itex]\frac{dF}{dx}[/itex] in my terms.

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

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:
[tex]\frac{dh}{du}=\frac{\sqrt{1+x^{2}}}{1+(\sqrt{1+x^{2}}-1)^{2}}-F(x)\frac{1+x^{2}}{x}[/tex]

Calin

[itex]\\int_1^4 \\sqrt{t}(ln(t))dt[/itex]

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.

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:
[tex]\frac{dF}{df}=\frac{\frac{dF}{dx}}{\frac{df}{dx}}[/tex]