Find f'''(1) for f(x) = x/2 sqrt(2-x²)+ arcsin x/sqrt2

Find $f'''(1)$ for $f(x) = \dfrac{x}{2}\sqrt{2-x^{2}}+\arcsin\dfrac{x}{\sqrt2}$.

Step 1: Differentiate the first term

Product rule: \[\frac{d}{dx}\left[\frac{x}{2}\sqrt{2-x^{2}}\right] = \frac12\sqrt{2-x^{2}}+\frac{x}{2}\cdot\frac{-x}{\sqrt{2-x^{2}}} = \frac{\bigl(2-x^{2}\bigr)-x^{2}}{2\sqrt{2-x^{2}}} = \frac{1-x^{2}}{\sqrt{2-x^{2}}}.\]

Step 2: Differentiate the second term

\[ \frac{d}{dx}\arcsin\frac{x}{\sqrt2} = \frac{1}{\sqrt{1-\dfrac{x^{2}}{2}}}\cdot\frac{1}{\sqrt2} = \frac{1}{\sqrt{2-x^{2}}}. \]

Step 3: Add: everything collapses

\[ f'(x) = \frac{1-x^{2}}{\sqrt{2-x^{2}}}+\frac{1}{\sqrt{2-x^{2}}} = \frac{2-x^{2}}{\sqrt{2-x^{2}}} = \sqrt{2-x^{2}}. \tag{1} \]

This is the whole point of the problem: the awkward expression is an antiderivative of $\sqrt{2-x^{2}}$, so its derivative is that square root and nothing more.

Step 4: Second derivative

\[ f''(x) = \frac12\bigl(2-x^{2}\bigr)^{-1/2}\cdot\bigl(-2x\bigr) = \frac{-x}{\sqrt{2-x^{2}}}. \]

Step 5: Third derivative

Product rule on $-x\bigl(2-x^{2}\bigr)^{-1/2}$: \[f'''(x) = -\bigl(2-x^{2}\bigr)^{-1/2}-x^{2}\bigl(2-x^{2}\bigr)^{-3/2} = \frac{-\bigl(2-x^{2}\bigr)-x^{2}}{\bigl(2-x^{2}\bigr)^{3/2}} = \frac{-2}{\bigl(2-x^{2}\bigr)^{3/2}}.\]

Step 6: Evaluate at $x = 1$

\[ 2-1^{2} = 1 \;\Longrightarrow\; f'''(1) = \frac{-2}{1^{3/2}} = -2. \]

Step 7: Verification

\[\int\sqrt{a^{2}-x^{2}}\,dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\arcsin\frac{x}{a}+C, \quad\text{here } a^{2} = 2 \;\;\checkmark\]

\[\text{so } f \text{ is that antiderivative and (1) is confirmed: } f' = \sqrt{2-x^{2}} \;\;\checkmark\]

\[f''' \lt 0 \text{ on the whole domain } |x|\lt \sqrt2, \text{ with } f'''\to-\infty \text{ at the endpoints} \;\;\checkmark\]