Derivative of a complex function $y=(\tan{2x})^{\cot{(\frac{x}{2})}}$

Question

If I replaced $y=(\tan{2x})^{\cot{(\frac{x}{2})}}$ by $y=e^{(\ln{(\tan{2x})}) \cdot (\cot(\frac{x}{2}))}$ and calculated $$y'=e^{(\ln{(\tan2x))} \cdot (\cot(\frac{x}{2}))} \cdot \left( \frac{2}{\frac{\cos^{2}(2x)}{\tan(2x)}} \cdot \cot(\frac{x}{2}) + (\ln{(\tan2x)}) \cdot (-\frac{\frac{1}{2}}{\sin^{2}(\frac{x}{2})}) \right).$$

Is it possible to further calculate the derivative of this complex function, or simplify it, or is it the final answer?

And yes, I know about $\csc$ and $\sec$, but I better get along without them.

Answer 1

We have

$$y=(\tan{2x})^{\cot{\frac{x}{2}}}=e^{\cot{\frac{x}{2}}\cdot\log (\tan{2x})}$$

and thus

$$y'=e^{\cot{\frac{x}{2}}\cdot\log (\tan{2x})}\cdot\left(-\frac1{2\sin^2\frac x 2}\cdot\log(\tan 2x)+\cot \frac x 2\frac{\frac 2 {\cos^2 2x}}{\tan 2x}\right)=$$

$$=(\tan{2x})^{\cot{\frac{x}{2}}}\left(\frac{4\cot \frac x 2}{\sin 4x}-\frac{\log(\tan 2x)}{2\sin^2\frac x 2}\right)$$

Answer 2

Well, it's not going to be super-short, but it definitely can be simplified quite a bit. Before we start simplifying, though, I should point out that you have an error in your expression for $y'$, which could be an actually error while finding the derivative or an error in typesetting it: the part $\frac{2}{\frac{\cos^2(2x)}{\tan(2x)}}$ must in fact be $\frac{2}{\cos^2(2x)\tan(2x)}$.

In fact, this is one of the issues here: you shouldn't keep "multi-level" fractions, but you should rather simplify them — after all, fractions presumably consist of a numerator and a denominator and nothing else, don't they? So by the same token, $\frac{\frac{1}{2}}{\sin^2(\frac{x}{2})}$ is the same as $\frac{1}{2\sin^2(\frac{x}{2})}$.

One more thing you should simplify is the first part of this derivative: that exponential function $e^{\cdots}$ is the same as the original function (remember how you started your solution?), so it will look a lot shorter if you go back to that original form for this part.

After that, there may be a few more things here and there to clean it up.

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As a separate note, let's look at these kinds of derivatives in general. Let $y=[f(x)]^{g(x)}$. Rewriting it as $y=[f(x)]^{g(x)}=e^{\ln(f(x))\cdot{g(x)}}$ and following the same procedure as you did for differentiating it, we'll come to $$\begin{multline*}y'=e^{\ln(f(x))\cdot{g(x)}}\cdot\left(\frac{f'(x)}{f(x)}\cdot g(x)+\ln(f(x))\cdot g'(x)\right)\\ =[f(x)]^{g(x)}\cdot\left(\frac{f'(x)}{f(x)}\cdot g(x)+\ln(f(x))\cdot g'(x)\right)\\ =g(x)\cdot f(x)^{g(x)-1}\cdot f'(x)+[f(x)]^{g(x)}\cdot\ln f(x)\cdot g'(x).\end{multline*}$$ So, interestingly enough, the correct derivative of $[f(x)]^{g(x)}$ can be found by differentiating it as a power function and as an exponential function and adding the two together!

Answer 3

$y = (\tan 2x)^{\cot \frac x2}\\ y = e^{(\cot \frac x2)\ln(\tan 2x)}\\ y' = e^{(\cot \frac x2)\ln(\tan 2x)}(-\frac 12\csc^2 \frac {x}{2}\ln\tan 2x + \frac {\cot \frac x2}{\tan 2x}2\sec^2 2x)\\ \frac {\sec 2x}{\tan 2x} = \csc 2x\\ y' = ((\tan 2x)^{\cot \frac x2})(-\frac 12\csc^2 \frac {x}{2}\ln\tan 2x + 2\cot \frac x2\csc 2x\sec 2x)\\ $

Now I would probably be inclined to leave it like that.

But you could say $\csc^2 \frac 12 x = \frac{2}{1-\cos x}$ and $\cot \frac 12 x = \frac {\sin x}{1-\cos x}$

$y' = ((\tan 2x)^{\cot \frac x2})(\frac {-\ln\tan 2x + 2\sin x\csc 2x\sec 2x}{1-\cos x})\\ 2\sin x \csc 2x = \frac {2\sin x}{2\sin x\cos x} = \sec x\\ y' = ((\tan 2x)^{\cot \frac x2})(\frac {-\ln\tan 2x + \sec x\sec 2x}{1-\cos x})\\$

Answer 4

For this kind of problem, I think that logarithmic differentiation mkes life easier $$y=\tan (2 x)^{\cot \left(\frac{x}{2}\right)}\implies \log(y)=\cot \left(\frac{x}{2}\right) \log (\tan (2 x))$$ Differentiating both sides $$\frac {y'}y=2 \cot \left(\frac{x}{2}\right) \csc (2 x) \sec (2 x)-\frac{1}{2} \csc ^2\left(\frac{x}{2}\right) \log (\tan (2 x))$$ which can be simplified $$\frac {y'}y=\frac{\log (\tan (2 x))-\sec (x) \sec (2 x)}{\cos (x)-1}$$ Now $$y'=y \times \left(\frac {y'}y \right)$$