Find y' for y = sqrt(1+ sinh²4x)
Find $y'$ for $y = \sqrt{1+\sinh^{2}4x}$.
Explanation of Method
A radical that disappears entirely under a hyperbolic identity.
Step 1: Simplify with the hyperbolic identity
From $\cosh^{2}t - \sinh^{2}t = 1$ we get $1+\sinh^{2}t = \cosh^{2}t$. With $t = 4x$: \[y = \sqrt{\cosh^{2}4x} = \left|\cosh 4x\right| = \cosh 4x,\]
the absolute value being harmless because $\cosh$ is positive everywhere.
Step 2: Differentiate
Chain rule with inner function $4x$: $y' = \sinh 4x\cdot 4 = 4\sinh 4x$.
Step 3: Verification
\[x = 0:\quad y = \sqrt{1} = 1,\quad y' = 4\sinh 0 = 0, \text{ the minimum of } y \;\;\checkmark\]
\[x = 0.25:\quad y' = 4\sinh 1 \approx 4.70081; \qquad \frac{y(0.25+10^{-5})-y(0.25-10^{-5})}{2\times 10^{-5}} \approx 4.70081 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = 4\sinh 4x\,}\]
Remark
Remark. Differentiating the radical directly would give $\dfrac{4\sinh 4x\cosh 4x}{\sqrt{1+\sinh^{2}4x}}$, which simplifies to the same answer — but recognising the identity first turns three steps into one. Note that the circular analogue $\sqrt{1+\sin^{2}4x}$ does not simplify, since $\sin^{2}+\cos^{2} = 1$ has a plus sign between the squares.