Differentiate 2(tan⁡x)ᶜᵒᵗ⁡ˣ with respect to x

To differentiate $y=2{(tanx)}^{cotx}$ with respect to $x$, we state the correct option clearly:
The correct option is 2 csc²x.tanxᶜᵒᵗˣ (1-log(tanx)).

Let $u={(tanx)}^{cotx}$. The original function becomes $y=2u$, which gives:
$$\frac{dy}{dx}=2\frac{du}{dx}$$

Taking the natural logarithm on both sides of the equation for $u$:
$$logu=log\left({(tanx)}^{cotx}\right)$$

Using the power property of logarithms, we simplify the right-hand side:
$$logu=cotx·log(tanx)$$

Differentiating both sides with respect to $x$ using the product rule:
$$\frac{1}{u}\frac{du}{dx}=\frac{d}{dx}(cotx)·log(tanx)+cotx·\frac{d}{dx}[log(tanx)]$$

Recall the differentiation formulas:
$\frac{d}{dx}(cotx)=-{csc}^{2}x$
and by using the chain rule:
$$\frac{d}{dx}[log(tanx)]=\frac{1}{tanx}·{sec}^{2}x=\frac{cosx}{sinx}·\frac{1}{{cos}^{2}x}=\frac{1}{sinxcosx}$$

Substitute these derivatives back:
$$\frac{1}{u}\frac{du}{dx}=-{csc}^{2}xlog(tanx)+cotx·\frac{1}{sinxcosx}$$

Simplifying the second term:
$$cotx·\frac{1}{sinxcosx}=\frac{cosx}{sinx}·\frac{1}{sinxcosx}=\frac{1}{{sin}^{2}x}={csc}^{2}x$$

Now we combine the terms:
$$\frac{1}{u}\frac{du}{dx}=-{csc}^{2}xlog(tanx)+{csc}^{2}x$$

Factoring out ${csc}^{2}x$:
$$\frac{1}{u}\frac{du}{dx}={csc}^{2}x(1-log(tanx))$$

Multiplying by $u$:
$$\frac{du}{dx}=u·{csc}^{2}x(1-log(tanx))$$

Substitute $u={(tanx)}^{cotx}$ back:
$$\frac{du}{dx}={(tanx)}^{cotx}{csc}^{2}x(1-log(tanx))$$

Multiplying by $2$ to get the final derivative of $y$:
$$\frac{dy}{dx}=2{csc}^{2}x·{(tanx)}^{cotx}(1-log(tanx))$$