## derivative

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??
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 Recognitions: Gold Member Homework Help Science Advisor 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 $$\frac{dx}{du}=\frac{1}{\cos^{2}u}=1+\tan^{2}u=1+x^{2}$$
 Recognitions: Gold Member Science Advisor 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 $\frac{df}{du}\frac{du}{dx} in that is the "chain" rule. ## derivative sotty,for threat 2 i don't really understand.can u plsexplain more clearly??  Recognitions: Gold Member Homework Help Science Advisor HallsofIvy's "u" is not the same as my "u". He is presenting a technique to evaluate the term [itex]\frac{dF}{dx}$ 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 $\frac{dh}{du}$
 can u pls show me the whole steps ??thanx
 Recognitions: Gold Member Homework Help Science Advisor Darn, I made a mistake! the correct expression for the derivative with respect to u=arctan(x), should be: $$\frac{dh}{du}=\frac{\sqrt{1+x^{2}}}{1+(\sqrt{1+x^{2}}-1)^{2}}-F(x)\frac{1+x^{2}}{x}$$
 $\\int_1^4 \\sqrt{t}(ln(t))dt$

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 Quote by teng125 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.
 Recognitions: Gold Member Homework Help Science Advisor 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: $$\frac{dF}{df}=\frac{\frac{dF}{dx}}{\frac{df}{dx}}$$