Differentiate 9ᵗᵃⁿ⁡³ˣ with respect to x

The correct option is: 9^tan(3x) (3 log(9) sec²(3x)).

To differentiate the given function with respect to $x$, we use the chain rule of differentiation.
Let $y=9^{tan\left(3x\right)}$.

Recall the standard differentiation formula for exponential functions:
$\frac{d}{du}\left(a^u\right)=a^{u}log\left(a\right)$

Applying the chain rule, we differentiate the outer exponential function first:
$\frac{dy}{dx}=9^{tan\left(3x\right)}log\left(9\right)·\frac{d}{dx}(tan(3x\left)\right)$

Next, we differentiate the trigonometric function $tan\left(3x\right)$ with respect to $x$, which also requires the chain rule:
$\frac{d}{dx}(tan(3x\left)\right)={sec}^2\left(3x\right)·\frac{d}{dx}\left(3x\right)$
Since $\frac{d}{dx}\left(3x\right)=3$, we have:
$\frac{d}{dx}(tan(3x\left)\right)=3{sec}^2\left(3x\right)$

Substituting this back into our primary derivative equation gives:
$\frac{dy}{dx}=9^{tan\left(3x\right)}log\left(9\right)·3{sec}^2\left(3x\right)$

Rearranging the terms to match the correct option layout:
$\frac{dy}{dx}=9^{tan\left(3x\right)}(3log(9\left){sec}^2\right(3x\left)\right)$

Note: The option text "sec2x" is a simplified shorthand notation representing the term ${sec}^2\left(3x\right)$ based on the functional argument.