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12.12 Derivative

  1. \(g(x)=|x|=\begin{cases}x,x\geq0\\-x,x<0\end{cases}\), find the \(g'(x)\)​

    Let's analyze when \(x>0\), then \(g^{\prime}(x)=\lim_{h\to0}\frac{\left|x+h\right|-\left|x\right|}{h}=\lim_{h\to0}\frac{x+h-x}{h}=1\)​

    When \(x<0\), then \(g^{\prime}(x)=\lim_{h\to0}\frac{|x+h|-|x|}{h}=\lim_{h\to0}\frac{-(x+h)-(-x)}{h}=\lim_{h\to0}\frac{-h}{h}=-1\)

    When \(x = 0\), then \(g'(0) = \lim_{h \to 0} \frac{|0 + h| - |0|}{h} = \lim_{h \to 0} \frac{|h|}{h}\)

    Then \(\lim_{h\to0^{+}}\frac{|h|}{h}=1\) but \(\lim_{h\to0^{-}}\frac{|h|}{h}=-1\), thus \(\lim_{h \to 0} \frac{|0 + h| - |0|}{h} \text{ does not exist.}\)

    Thus \(g\) is not differentiable at \(x=0\)

    Therefore \(g^{\prime}(x)=\begin{cases}1,x>0\\ -1,x<0\end{cases}\)​

    Note: This also verify the theorem: if \(f\) is differentiable, then \(f\) is continuous

  2. \(f(x) =\begin{cases}x^2 \cdot \sin\left(\frac{1}{x}\right), & \text{if } x \neq 0 \\0, & \text{if } x = 0\end{cases}\)

    1. Is \(f\) differentiable at \(x = 0\)?

      \(f'(0) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \cdot \sin\left(\frac{1}{h}\right) - 0}{h} = \lim_{h \to 0} h \cdot \sin\left(\frac{1}{h}\right)=0\)

      Thus: \(f'(0) = 0\) 2. Is \(f\) continuous at \(x = 0\)?

      By the theorem, since \(f'(0)\) exists, \(f\) is continuous at \(x = 0\).

  3. \(h(x)=x^{x}=e^{x\cdot\ln x}=e^{x\cdot\ln x}\left(x\ln x\right)^{\prime}=x^{x}\left(\ln x+1\right)\)​

\(\sin: \mathbb{R} \to [-1, 1]\)

  • \(\sin x\) is continuous.
  • \(\sin x\) is a periodic function:\(\sin(x) = \sin(x + 2\pi), \forall x\)

  • \(\sin(x) = -\sin(-x)\)

  • \(\sin(x)\) is not injective.

Consider

\(\sin: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \to [-1, 1]\) is injective and surjective.

The inverse is \(\sin^{-1}: [-1, 1] \to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)

We denote \(\sin^{-1}(t) = \arcsin(t)\) which is \(\arcsin(t) = s \iff \sin(s) = t\)

Then \(\arcsin(t)\) is continuous in \([-1, 1]\).

Thus we have \(\arcsin(\sin(x)) = x, \quad -\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\) and \(\sin(\arcsin(t)) = t, \quad -1 \leq t \leq 1\)

From Lecture Class:
If \(f\) is differentiable at \(x_0\) and \(f'(x_0) \neq 0\), then \(f^{-1}\) is differentiable in \(y_0 = f(x_0)\).

\((f^{-1})'(y_0) = \frac{1}{f'(f^{-1}(y_0))}\)

\(f(x) = \sin(x)\) \(-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\), \(f'(x) = \cos(x)\) \(f'(x) = \cos(x) \neq 0\)

By theorem, \(\arcsin(t)\) is differentiable and \((\arcsin(t))' = \frac{1}{\cos(\arcsin(t))}=\frac{1}{\sqrt{1-t^2}}\)

Unit circle: image​

\(a^2 + b^2 = 1\), \(\cos(s) = \frac{a}{1} = a, \quad \sin(s) = \frac{b}{1} = b\), \(\cos^2(s) + \sin^2(s) = 1\)

Then \(\cos^2(s) = 1 - \sin^2(s)\) \(\Rightarrow\) \(\cos(s)=\sqrt{1-\sin^2(s)}=\sqrt{1-t^2}\) since \(\cos(s) > 0\)

where \(s = \arcsin(t)\) and \(\sin(s) = \sin(\arcsin(t)) = t\)