$f$ bounded but $f'$ isn't

complex-analysis analysis examples-counterexamples

Solution 1:

$(z+1)\log(z+1)$ is bounded on the open unit disk $|z|\lt1$. Its derivative, $1+\log(z+1)$ is not.

Solution 2:

Any branch of $\sqrt{1+z}$ is bounded in the unit disk, but the derivative $\frac{1}{2\sqrt{1+z}}$ is not.

Related

Prove that $(n - 1)^2 \mid n^k -1$ if and only if $(n - 1) \mid k$ [duplicate]
Trigonometric Equation $\sin x=\tan\frac{\pi}{15}\tan\frac{4\pi}{15}\tan\frac{3\pi}{10}\tan\frac{6\pi}{15}$
Prove: $\sum _{n=1}^{\infty} \left(\frac{n^{2} + 1}{n^{2} +n +1}\right)^{n^2} < \infty$ without L'Hôpital's
Calculate coordinates of 3rd point (vertex) of a scalene triangle if angles and sides are known.
Increasing function $ f : \mathbb R ^ + \to \mathbb R $ with $ x f ( x ) + 2 > 0 $ and $ f ( x ) f \left( f ( x ) + \frac 4 x \right) = 1 $
So-called Artin-Schreier Extension
Multiplicative inverse of $2x + 3 + I$ in $\mathbb{Z}_5[x]/I $?
The coupon collectors problem [closed]
show that the ordered square is locally connected~
Monotone functions and continuity
Why there is no continuous argument function on $\mathbb{C}\setminus\{0\}$?
How far is being star compact from being countably compact？