Find y' for y = cosh x - x sinh x
Find $y'$ for $y = \cosh x - x\sinh x$.
Explanation of Method
A difference in which two terms cancel exactly.
Step 1: Recall the standard hyperbolic derivatives
\[ \bigl(\cosh x\bigr)' = \sinh x, \qquad \bigl(\sinh x\bigr)' = \cosh x. \]
Unlike the circular case, no minus sign appears in either rule.
Step 2: Differentiate the first term
\[ \frac{d}{dx}\bigl(\cosh x\bigr) = \sinh x. \]
Step 3: Differentiate the second term by the product rule
With $u = x$, $u' = 1$, $v = \sinh x$, $v' = \cosh x$: $\frac{d}{dx}\bigl(x\sinh x\bigr) = \sinh x + x\cosh x$.
Step 4: Combine
The two $\sinh x$ terms cancel: $y' = \sinh x - \bigl(\sinh x + x\cosh x\bigr) = -x\cosh x$.
Step 5: Verification
\[x = 0:\quad y(0) = \cosh 0 - 0 = 1,\quad y'(0) = 0, \text{ the maximum of } y \;\;\checkmark\]
\[x = 1:\quad y' = -\cosh 1 \approx -1.5431; \qquad \frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx -1.5431 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = -x\cosh x\,}\]
Remark
Remark. Because $\cosh x \gt 0$ always, the sign of $y'$ is the sign of $-x$: the function increases on $(-\infty,0)$ and decreases on $(0,\infty)$, so $y(0) = 1$ is a global maximum. Read backwards, the result says $\cosh x - x\sinh x$ is an antiderivative of $-x\cosh x$.