Find y' for y = arsinh x · arcosh x
Find $y'$ for $y = \operatorname{arsinh} x \cdot \operatorname{arcosh} x$.
Explanation of Method
A product of two inverse hyperbolic functions.
Step 1: Recall the standard derivatives
\[ \bigl(\operatorname{arsinh} x\bigr)' = \frac{1}{\sqrt{x^{2}+1}}, \qquad \bigl(\operatorname{arcosh} x\bigr)' = \frac{1}{\sqrt{x^{2}-1}}. \]
The first holds for every real $x$; the second requires $x \gt 1$, which therefore fixes the domain of the product.
Step 2: Identify the factors and apply the product rule
For $y = uv$ recall $y' = u'v + uv'$. Here
\[ u = \operatorname{arsinh} x,\quad u' = \frac{1}{\sqrt{x^{2}+1}}, \qquad v = \operatorname{arcosh} x,\quad v' = \frac{1}{\sqrt{x^{2}-1}}. \]
Step 3: Substitute
\[ y' = \frac{\operatorname{arcosh} x}{\sqrt{x^{2}+1}} + \frac{\operatorname{arsinh} x}{\sqrt{x^{2}-1}}. \]
Step 4: Verification
\[x = 2:\quad \operatorname{arsinh} 2 \approx 1.4436,\ \operatorname{arcosh} 2 \approx 1.3170\]
\[y' \approx \frac{1.3170}{\sqrt5} + \frac{1.4436}{\sqrt3} \approx 0.5890 + 0.8335 = 1.4225\]
\[\frac{y(2+10^{-5})-y(2-10^{-5})}{2\times 10^{-5}} \approx 1.4225 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{\operatorname{arcosh} x}{\sqrt{x^{2}+1}} + \dfrac{\operatorname{arsinh} x}{\sqrt{x^{2}-1}}, \qquad x > 1\,}\]
Remark
Remark. Both terms are positive on $x \gt 1$, so $y$ is strictly increasing there. As $x \to 1^{+}$ the second term blows up (because $\sqrt{x^{2}-1} \to 0$ while $\operatorname{arsinh} 1 \approx 0.881 \neq 0$), so the graph starts with a vertical tangent at $x = 1$, where $y = 0$. In logarithmic form $\operatorname{arsinh} x = \ln\bigl(x+\sqrt{x^{2}+1}\bigr)$ and $\operatorname{arcosh} x = \ln\bigl(x+\sqrt{x^{2}-1}\bigr)$.