Differentiate (cos⁡3x)³ˣ with respect to x

The correct option is:
(cos3x)³ˣ (3 log(cos3x) – 9x tan3x)

To differentiate the given function with respect to $x$, we represent the function as:
$y={(cos3x)}^{3x}$

Since the variable $x$ appears in both the base and the exponent, we use logarithmic differentiation. Taking the natural logarithm ($log$) on both sides of the equation:
$logy=log\left[{(cos3x)}^{3x}\right]$

Using the logarithmic property $log\left(a^b\right)=bloga$, we can rewrite the equation as:
$logy=3xlog(cos3x)$

Now, we differentiate both sides with respect to $x$.
On the left-hand side, applying the chain rule, we get:
$\frac{d}{dx}(logy)=\frac{1}{y}\frac{dy}{dx}$

On the right-hand side, we apply the product rule of differentiation: $\frac{d}{dx}\left(uv\right)=u\frac{dv}{dx}+v\frac{du}{dx}$, where $u=3x$ and $v=log(cos3x)$:
$\frac{d}{dx}[3xlog(cos3x)]=3x·\frac{d}{dx}[log(cos3x)]+log(cos3x)·\frac{d}{dx}\left(3x\right)$

Let us calculate each derivative:
1. Differentiating $3x$ gives:
$\frac{d}{dx}\left(3x\right)=3$
2. Differentiating $log(cos3x)$ using the chain rule gives:
$\frac{d}{dx}[log(cos3x)]=\frac{1}{cos3x}·\frac{d}{dx}(cos3x)$
$=\frac{1}{cos3x}·(-sin3x)·\frac{d}{dx}\left(3x\right)$
$=\frac{-sin3x}{cos3x}·3=-3tan3x$

Substituting these results back into the product rule equation:
$\frac{1}{y}\frac{dy}{dx}=3x(-3tan3x)+3log(cos3x)$
$\frac{1}{y}\frac{dy}{dx}=3log(cos3x)-9xtan3x$

Multiplying both sides by $y$ to solve for $\frac{dy}{dx}$:
$\frac{dy}{dx}=y[3log(cos3x)-9xtan3x]$

Substituting $y={(cos3x)}^{3x}$ back into the expression:
$\frac{dy}{dx}={(cos3x)}^{3x}(3log(cos3x)-9xtan3x)$