Find f'(1) for f(x) = ln(1+x) + arcsin x/2
Find $f'(1)$ for $f(x) = \ln(1+x) + \arcsin\dfrac{x}{2}$.
Explanation of Method
A logarithm plus an arcsine, evaluated at a point.
Step 1: Differentiate term by term
The first term is standard; the second needs the chain rule with $u = \dfrac{x}{2}$, $u' = \dfrac12$: $f'(x) = \frac{1}{1+x} + \frac{1}{\sqrt{1-\dfrac{x^{2}}{4}}}\cdot\frac{1}{2}$.
Step 2: Simplify the second term
Pull the $\tfrac14$ out of the radical: $\sqrt{1-\dfrac{x^{2}}{4}} = \dfrac{\sqrt{4-x^{2}}}{2}$, so the factor $\tfrac12$ cancels against it: \[\frac{1}{2}\cdot\frac{2}{\sqrt{4-x^{2}}} = \frac{1}{\sqrt{4-x^{2}}}, \qquad\text{hence}\qquad f'(x) = \frac{1}{1+x} + \frac{1}{\sqrt{4-x^{2}}}.\]
Step 3: Evaluate at $x = 1$
\[ f'(1) = \frac{1}{2} + \frac{1}{\sqrt{4-1}} = \frac{1}{2} + \frac{1}{\sqrt3}. \]
Step 4: Verification
\[\frac{f(1+10^{-5})-f(1-10^{-5})}{2\times 10^{-5}} \approx 1.07735 \;\;\checkmark\]
\[\text{Common denominator: } \frac12+\frac{\sqrt3}{3} = \frac{3}{6}+\frac{2\sqrt3}{6} = \frac{3+2\sqrt3}{6} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,f'(1) = \dfrac{1}{2} + \dfrac{1}{\sqrt3} = \dfrac{1}{2} + \dfrac{\sqrt3}{3} = \dfrac{3+2\sqrt3}{6} \approx 1.07735\,}\]
Remark
Remark. The domain is $-1 \lt x \leq 2$ for $f$ itself, but the derivative additionally needs $|x| \lt 2$, so the formula holds on $(-1,2)$. The point $x = 1$ sits comfortably inside. Both terms are positive there, so $f$ is increasing.