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Find the derivative of $2^{\tan(1/x)}$. I know that I should replace $\frac1x$ with $u$ and such, but then I can't continue it...

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Derivative with respect to what? –  Cameron Buie Nov 14 '12 at 6:17
    
Oh sorry,it is 2^tg(1/x) –  xdfg Nov 14 '12 at 6:27
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@ctype.h: If you are going to bump lots of old questions to the front page merely to fix the grammar, could you please limit this to about five edits a day or so? See this meta discussion. Right now I cannot tell new questions from old ones on the front page. –  Rahul Nov 22 '12 at 3:56
    
@RahulNarain Sorry, I was not aware that it would be problematic. I will scale down the editing. –  ctype.h Nov 22 '12 at 6:52
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1 Answer

Hint: $2^{\tan(1/x)}=e^{(\ln 2)(\tan(1/x)}$.

If $y=e^u$, then by the Chain Rule, $\dfrac{dy}{dx}=e^u \dfrac{du}{dx}$.

Now let $u=(\ln 2)\tan(1/x)$. I think you know how to find $\dfrac{du}{dx}$.

Another way: Let $y=2^{\tan(1/x)}$. Take the natural logarithm of both side. We get $$\ln y=(\ln 2)\tan(1/x).$$ Now differentiate both sides with respect to $x$. On the left we get $\dfrac{1}{y}\dfrac{dy}{dx}$.

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the answer in my textbook is [-ln2*2^tg3/2]/x^2*cos^2 x.. but I need this now,can you post the whole solution PLEASE? –  xdfg Nov 14 '12 at 6:56
    
For the first way I suggested doing it, we have $e^u=2^{\tan(1/x)}$. But $u=(\ln 2)\tan(1/x)$. To differentiate $(\ln 2)\tan(1/x)$, that is, to find $\frac{du}{dx}$, use the Chain Rule, again. Let $v=1/x$. We get $(\ln 2)(-1/x^2)\sec^2(1/x)$. To get the form you were given, note that $\sec^2 x=\frac{1}{\cos^2 x}$. I imagine you can put the pieces together. The fact that $\frac{d}{dv}(\tan v)=\sec^2 v=\frac{1}{\cos^2 v}$ is an I hope familiar rule of differentiation. If you don't know it, you can get it by differentiating $\frac{\sin v}{\cos v}$ using the Quotient Rule. –  André Nicolas Nov 14 '12 at 7:09
    
Why does the typesetting look weird in the exponents in your post despite you using \ before $\tan$? –  Joe Nov 22 '12 at 3:57
    
@Joe: Don't know. Small fractions render funny (with real LaTeX they look OK). –  André Nicolas Nov 22 '12 at 4:05
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