Calculating the value of $f'(x)$ if it exists.

Question

Note that I am not looking for a solution, just a general advice on how to approach these questions and practical tips.

We've just started differentiability and I'm completely lost. I'm asked to calculate the value of $f'(x)$ at $x_0$ or prove it doesn't exist. But we haven't done much practice and only focused on theorems so far so I'm not sure how to approach these questions.

Are there any special tricks regarding calculating differential function limits or is it basically the same as ordinary limits?

$f\left(x\right)\ =\ \left|x^{2}-3x-4\right|,\ \ \ x_{0}=4$

$\lim _{x\to 4}\frac{\left(f\left(x\right)-f\left(4\right)\right)}{x-4} =\ \lim _{x\to 4}\frac{\left|x^2-3x-4\right|-0}{x-4}$

$\lim _{x\to 4^+}\frac{x^2-3x-4}{x-4} =\ \lim _{x\to 4^+}\frac{\left(x-4\right)\left(x-3\right)+4x-16}{x-4} =\ \lim _{x\to 4^+}\frac{\left(x-4\right)\left(x-3\right)+4\left(x-4\right)}{x-4} $

$=\ \lim _{x\to 4^+}\left(x-3\right)+4 =\ \lim _{x\to 4^+}\left(4-3\right)+4 = 5$

$\lim _{x\to 4^-}\frac{-x^2+3x+4}{x-4} =\ \lim _{x\to 4^-}\frac{\left(x-4\right)\left(3-x\right)-4x+16}{x-4} =\ \lim _{x\to 4^-}\frac{\left(x-4\right)\left(3-x\right)-4\left(x-4\right)}{x-4} $

$=\ \lim _{x\to 4^-}\left(3-x\right)-4 = \left(3-4\right)-4 = -5$

Therefore the limit doesn't exists, and f is not differentiable at 4.

b.

$f\left(x\right)\ =\ \sqrt[3]{x^{2}-\left|x\right|},\ \ x_{0}\ =0$

$\lim \:_{x\to \:0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x-0} =\ \lim \:_{x\to \:0}\frac{\left(\sqrt[3]{x^2-x}-0\right)}{x}$

$\lim _{x\to 0^+}\frac{\sqrt[3]{x^2-x}}{x}\cdot \frac{\sqrt[\frac{2}{3}]{x^2-x}}{\sqrt[\frac{2}{3}]{x^2-\:x}}=\lim \:_{x\to 0^+}\frac{\left(x^2-x\right)}{x\left(\sqrt[\frac{2}{3}]{x^2-\:x}\right)}=\lim \:_{x\to \:0^+}\frac{x-1}{\sqrt[\frac{2}{3}]{x^2-\:x}}=\frac{0-1}{\sqrt[\frac{2}{3}]{0^+}}=-∞$

A one sided limit doesn't exist, therefore f is not diffirintable at 0.

c.

$f\left(x\right)= \begin{cases} \frac{\sin x^{2}}{x} & if \ x ≠ 0\\ 0 & if \ x = 0 \\ \end{cases} \ , \ x_{0}\ =\ 0$

$\lim \:\:_{x\to \:\:0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x-0} =\ \lim \:\:_{x\to \:\:0^{ }}\frac{f\left(x\right)}{x} $

$\lim _{x\to 0^+}\frac{\left(\frac{sin\left(x^2\right)}{x}\right)}{x}=\lim \:_{x\to 0^+}\left(\frac{sin\left(x^2\right)}{x^2}\right)=1$

$\lim _{x\to 0^-}\frac{\left(\frac{sin\left(x^2\right)}{x}\right)}{x}=\lim \:_{x\to 0^-}\left(\frac{sin\left(x^2\right)}{x^2}\right)=1$

Therefore the limit exists and f is differentiable at 0.

Answer 1

In both cases you can analyze the difference quotient $\frac{f(x)-f(x_0)}{x-x_0}$ which is the basis of the definition of the derivative. As $f(x_0)=f(0)=0$ in both cases, this simplifies to $$ -\frac{\sqrt[3]{|x|-x^2}}x=-\sqrt[3]{\frac1{x|x|}-\frac1x} $$ which diverges at $x=0$ and $$ \frac{\sin(x^2)}{x^2} $$ which has a well-known limit at $x=0$.

Answer 2

Do the first one by writing the left and right pieces as $f(x)=(x^2+x)^{1/3}, x<0; f(x)(x^2-x)^{1/3}, x \ge 0$ So the left derivative id $L f'(x)=\frac{1}{3}(x^2+x)^{-2/3}(2x+1).$ So $Lf'(0)=\infty$ so left derivative doe not exist, similarly right one also is infinite so it also does not exist. So the derivative does not exist at $x=0$

Second one you do by $h$-method: $$Lf'(0)=\lim_{ h \to 0} \frac{f(0-h)-f(0)}{-h}=\lim_{h \to 0}\frac{\frac{\sin^2 h}{- h}-0}{-h}=\lim_{h\to 0}(\frac{\sin h}{h})^2=1 $$ $$Rf'(0)=\lim_{ h \to 0} \frac{f(0+h)-f(0)}{h}=\lim_{h \to 0}\frac{\frac{\sin^2 h}{h}-0}{h}=\lim_{h\to 0}(\frac{\sin h}{h})^2=1 $$ Both being finite and equal $f(x)$ is differentiable at $x=0$.