<s>[INST] Power Inside Trigonometry Derivative with Chain Rule Trying to differentiate the following function with the chain rule, but I'm stumped with the $\cos(2x+3)^3$. Do I power the $(2x+3)$ by $3$?
$${\sinh ^2 (\cos(2x+3)^3)}$$
 [/INST] \begin{align*}
\frac{d}{dx} \left(\sinh^2 \left(\cos \left(2x+3\right)^3\right)\right)
&= 2 \sinh \left(\cos \left(2x+3\right)^3\right) \cdot \frac{d}{dx}\left(\sinh \left(\cos \left(2x+3\right)^3\right)\right)
\\
&= 2 \sinh \left(\cos \left(2x+3\right)^3\right) \cosh \left(\cos \left(2x+3\right)^3\right) \cdot \frac{d}{dx}\left(\cos \left(2x+3\right)^3\right).
\end{align*}
Can you continue from here?
As for the power, $\cos (2x+3)^3 = \cos \left((2x+3)^3\right)$, while $\cos^3 (2x+3) = \left(\cos (2x+3)\right)^3$.
As requested:
\begin{align*}
\frac{d}{dx}\left(\cos \left(2x+3\right)^3\right)
&= -\sin(2x+3)^3 \cdot \frac{d}{dx} \left((2x+3)^3\right)
\\
&= -\sin(2x+3)^3 \cdot 3(2x+3)^2 \cdot \frac{d}{dx}\left(2x+3\right)
\\
&= -\sin(2x+3)^3 \cdot 3(2x+3)^2 \cdot 2.
\end{align*}
</s>