Find y' for y = arccos e^x + e^(arcsin x)

Find $y'$ for $y = \arccos e^{x} + e^{\arcsin x}$.

Step 1: Differentiate $\arccos e^{x}$

Outer function $\arccos u$, inner $u = e^{x}$ with $u' = e^{x}$: \[\frac{d}{dx}\bigl(\arccos e^{x}\bigr) = -\frac{1}{\sqrt{1-u^{2}}}\cdot u' = -\frac{e^{x}}{\sqrt{1-e^{2x}}}.\]

Step 2: Differentiate $e^{\arcsin x}$

Here the composition is reversed: outer $e^{u}$, inner $u = \arcsin x$ with $u' = \dfrac{1}{\sqrt{1-x^{2}}}$: $\frac{d}{dx}\bigl(e^{\arcsin x}\bigr) = e^{u}\cdot u' = \frac{e^{\arcsin x}}{\sqrt{1-x^{2}}}$.

Step 3: Verification

\[x = -1:\quad e^{-1} \approx 0.36788 \;\Longrightarrow\; \text{first term} = -\frac{0.36788}{\sqrt{1-0.13534}} \approx -0.39562\]

\[\text{second term} = \frac{e^{\arcsin(-1)}}{\sqrt{1-1}} \quad\text{is undefined at } x = -1\]

\[x = -0.5:\quad y' \approx -0.76287 + 0.68403 \approx -0.07885 \;\;\checkmark\]

Step 4: Domain

Two separate restrictions apply. The first term needs $e^{2x} \lt 1$, i.e. $x \lt 0$; the second needs $|x| \lt 1$. Together the derivative is valid on $-1 \lt x \lt 0$.