Find y' for y = sinh³ x
Find $y'$ for $y = \sinh^{3} x$.
Explanation of Method
A cube of the hyperbolic sine, via the three-factor product rule.
Step 1: Write the cube as a product and apply the three-factor rule
With $y = \sinh x\cdot\sinh x\cdot\sinh x$ and $y' = u'vw + uv'w + uvw'$, where $u = v = w = \sinh x$ and $u' = v' = w' = \cosh x$: $y' = \cosh x\sinh^{2}x + \sinh x\cosh x\sinh x + \sinh^{2}x\cosh x$.
Step 2: Collect the three identical terms
\[ y' = 3\sinh^{2}x\cosh x. \]
Step 3: Verification
This matches the power-and-chain-rule pattern $\bigl(u^{n}\bigr)' = nu^{n-1}u'$ with $u = \sinh x$, $u' = \cosh x$, $n = 3$: $y' = 3\sinh^{2}x\cdot\cosh x \;\;\checkmark$.
\[x = 0:\quad \sinh 0 = 0 \;\Longrightarrow\; y'(0) = 0, \text{ a stationary inflection of } y \;\;\checkmark\]
\[x = 1:\quad y' = 3\sinh^{2}1\cosh 1 \approx 3(1.1752)^{2}(1.5431) \approx 6.393 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = 3\sinh^{2}x\cosh x\,}\]
Remark
Remark. Since $\cosh x \gt 0$ and $\sinh^{2}x \geq 0$, the derivative is non-negative everywhere: $\sinh^{3}x$ is increasing on $\mathbb{R}$, with a single stationary point at the origin that is not an extremum.