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

The correct option is: (3 cosx)^x (log(3 cosx) - x tanx).

To differentiate the function $y={\left(3cosx\right)}^x$ with respect to $x$, we can use the method of logarithmic differentiation. This method is particularly useful when a function has a variable in both the base and the exponent.

Let us define the function as:
$y={\left(3cosx\right)}^x$

Step 1: Take the natural logarithm (ln or log) of both sides.
Using the logarithmic property $log\left(a^b\right)=blog\left(a\right)$, we get:
$logy=log\left[{\left(3cosx\right)}^x\right]$
$logy=xlog\left(3cosx\right)$

Step 2: Differentiate both sides with respect to $x$.
We apply the chain rule on the left side and the product rule on the right side. The product rule states that $\frac{d}{dx}\left[u·v\right]=u\frac{dv}{dx}+v\frac{du}{dx}$, where $u=x$ and $v=log\left(3cosx\right)$.

Differentiating the left side:
$\frac{d}{dx}\left(logy\right)=\frac{1}{y}\frac{dy}{dx}$

Differentiating the right side:
$\frac{d}{dx}\left[xlog\left(3cosx\right)\right]=x·\frac{d}{dx}\left[log\left(3cosx\right)\right]+log\left(3cosx\right)·\frac{d}{dx}\left(x\right)$

Using the chain rule to differentiate $log\left(3cosx\right)$:
$\frac{d}{dx}\left[log\left(3cosx\right)\right]=\frac{1}{3cosx}·\frac{d}{dx}\left(3cosx\right)=\frac{1}{3cosx}·\left(-3sinx\right)=-\frac{3sinx}{3cosx}=-tanx$

Now, substitute this derivative and $\frac{d}{dx}\left(x\right)=1$ back into the product rule equation:
$\frac{1}{y}\frac{dy}{dx}=x\left(-tanx\right)+log\left(3cosx\right)·1$
$\frac{1}{y}\frac{dy}{dx}=log\left(3cosx\right)-xtanx$

Step 3: Solve for $\frac{dy}{dx}$ by multiplying both sides by $y$.
$\frac{dy}{dx}=y\left[log\left(3cosx\right)-xtanx\right]$

Finally, substitute the original expression for $y$ back into the equation:
$\frac{dy}{dx}={\left(3cosx\right)}^x\left[log\left(3cosx\right)-xtanx\right]$

This matches the correct option.