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 $3$?
$${\sinh ^2 (\cos(2x+3)^3)}$$
 A: \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*}
A: There is an alternative way, using the double angle formula for $\sinh^2 x$.
Let $w = \cos (2x+3)^3$. Using the identity $\sinh^2 x = \frac {1}{2} (\cosh 2x -1)$ and differentiating, we get $$\frac {1}{2} (\sinh 2w (2)) = \sinh 2w \ dw$$
Now we need to find $dw$, meaning we differentiate $\cos (2x+3)^3$.  Letting $z = 2x + 3$ and $dz = 2$ we have $(\cos^3 z) = -3 \cos^2 z (\sin z) (2)$ or $3 \sin 2z \cos z$ when we factor out $\cos z$ and see that $2 \sin z \cos z = \sin 2z$.
Putting all the pieces together we get
$$\dfrac {dy}{dx}{\sinh ^2 (\cos(2x+3)^3)} = -\sinh 2(2x+3)\cdot 3 \sin 2(2x+3) \cos (2x+3).$$
NOTE: If you didn't want to use the double angle formulas, you can differentiate $\sinh^2 w$ as $2 \sinh w \cosh w$ and (after differentiating the other parts) leave the answer in powers of cosine and sine.  The answer would then be $$\dfrac {dy}{dx}{\sinh ^2 (\cos(2x+3)^3)} = -2 \sinh (2x+3) \cosh (2x+3) \cdot 3 \sin (2x+3) \cos^2 (2x+3)$$
