[SOLVED] Help finding f(x)'.... THX!

**stickyMATH** · January 29th 2010, 07:01 PM

Can someone give me a hint with this problem?
The problem is:

(tan x)^(arctan x)

I was trying to use:
Dx (a)^u = (a)^u * ln(a) * Dx u

I am getting:
(tan)^(arctan x) * ln(tanx) * 1/(1+x^2)

The answer in the back of my book is:
(tan)^(arctan x) * [arctan(x)cot(x)sec^2(x) + ln(tanx)/(1+x^2)

I am messing up with the chain rule somewhere I think =(.

Again just a hint will do! Thanks!!!

**Sudharaka** · January 29th 2010, 07:10 PM

Dear stickyMATH,

$f(x)=(tanx)^{arctanx}$

$lnf(x)=arc(tanx).ln(tanx)$

Differenciating w.r.t x,

$\frac{d}{df(x)}lnf(x)\frac{df(x)}{dx}=ln(tanx)\fra c{d}{dx}(arctanx)+arctanx\frac{d}{dx}ln(tanx)$

$\frac{1}{f(x)}\frac{df(x)}{dx}=\frac{ln(tanx)}{1+x ^2}+(arctanx)cotxsec^{2}x$

$\frac{df(x)}{dx}=(tanx)^{arctan x}\left[(arctanx)cot(x)sec^2(x) + \frac{ln(tanx)}{(1+x^2)}\right]<br />$

Hope this helps.

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