Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question
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
@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

1 Answer 1

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 $\dfrac{1}{y}\dfrac{dy}{dx}$.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.