Finding $\frac{d}{dx}(\sqrt{3\cos^3x})$ I am attempting to find the derivative of $\sqrt{3\cos^3 x}$.

Using the chain rule, I got:

$$\frac{1}{2} \sqrt{(3\cos x)^3} × 3(3\cos x)^2 × (-3\sin x).$$

Apparently this is where I went wrong, the correct answer stated that instead of the derivative of the inner function should have been $-\sin x$ instead of $-3\sin x$. To my understanding the derivative of $3\cos x$ is $-3\sin x$. Is my reasoning incorrect?
$$\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x}\sqrt{3\cos^3x} &amp;= \sqrt{3}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\cos^\frac32x\\
&amp;= \sqrt{3}\cdot\frac32\cos^{\frac32 - 1}x\cdot\frac{\mathrm{d}}{\mathrm{d}x}\cos x\\
&amp;= \sqrt{3}\cdot\frac32\cos^\frac12x\cdot(-\sin x) \\
&amp;= -\frac{3^\frac32}{2}\sin x\sqrt{\cos x}
\end{align*}$$
Here is a correct implementation of chain rule

$\frac{d\sqrt{3cos^3(x)}}{dx} = \frac{1}{2\sqrt{3cos^3(x)}}*\frac{d( {3cos^3(x)})}{dx}$ = $\frac{1}{2\sqrt{3cos^3(x)}}*3*3*cos^2(x)*\frac{d({cos(x)})}{dx}$

$\qquad \qquad  =  \frac{1}{2\sqrt{3cos^3(x)}}*9*cos^2(x)*(-sin(x)) = -\frac{3^{\frac{3}{2}}}{2}*sin(x)*\sqrt{cos(x)}$