Skip to content

12.13 Derivative

Function Definition

\(f(x) = \tan(x) = \frac{\sin(x)}{\cos(x)}\)

image​

Domain of \(\tan(x)\)

\(\text{Dom } \tan : \mathbb{R} - \left\{ \dots, -\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots \right\}\)

\(\text{Dom } \tan : \mathbb{R} - \left\{ \left( 2k+1 \right) \frac{\pi}{2} \;|\; k \in \mathbb{Z} \right\}\)

Properties of \(\tan(x)\)

  • \(\tan(x)\) is continuous in its domain.
  • \(\tan(x) = \tan(x + \pi)\): \(\tan(x)\) is periodic.
  • \(x=\frac{\left(2k+1\right)\pi}{2}\) are vertical asymptotes.

Derivative of \(\tan(x)\)

\(f^{\prime}(x)\)​\(=\lim_{h\to0}\frac{\tan(x+h)-\tan(x)}{h}\)

Since \(\tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}\), then \(f'(x) = \lim_{h \to 0} \frac{\frac{\tan(x) + \tan(h)}{1 - \tan(x)\tan(h)} - \tan(x)}{h}\)

\(f'(x) = \lim_{h \to 0} \frac{\tan(x) + \tan(h) - \tan(x) (1 - \tan(x) \tan(h))}{h (1 - \tan(x) \tan(h))}\)

\(f'(x) = \lim_{h \to 0} \frac{\tan(h)(1 + \tan^2(x))}{h(1 - \tan(x)\tan(h))}\)

\(f'(x) = \frac{1 + \tan^2(x)}{1}\)

\(f'(x) = 1 + \tan^2(x)\)

\(\tan(x)\) is not injective. Let's consider \(\tan : \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \to \mathbb{R}\) is injective and surjective.

Then the inverse \(\tan^{-1} : \mathbb{R} \to \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)\)​\(\text{ such that }\tan(x)=t\), then \(\arctan(t) := \tan^{-1}(t)\)

Properties:

\(\tan(\arctan(t)) = t, \quad t \in \mathbb{R}\)

\(\arctan(\tan(x)) = x, \quad -\frac{\pi}{2} < x < \frac{\pi}{2}\)

To apply the theorem of the inverse function, we need to check that:

  1. \(\tan(x)\) is continuous in \(\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)\).

Since \(\tan(x) = \frac{\sin(x)}{\cos(x)}\) is the quotient of two continuous functions and \(\cos(x) \neq 0\) in \(\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)\).

  1. \(f^{\prime}(x)=\left(\tan(x)\right)^{\prime}=1+\tan^2\left(x\right)\) is continuous in \(\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)\) (by algebra of continuous functions).

  2. \(f'(x) = 1 + \tan^2(x) \neq 0, \quad \forall x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)\).

Then by the theorem, \(\arctan(t)\) is differentiable and

\((\arctan(t))' = \frac{1}{1 + \tan^2(\arctan(t))} = \frac{1}{1 + t^2}\)​\(\tan^2(\arctan(t)) = (\tan(\arctan(t)))^2 = t^2\)

Inverse of Cosine

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

  • \(\cos(x)\) is continuous.
  • \(\cos(x) = \cos(-x)\): \(\cos(x)\) is an even function.
  • \(\cos(x) = \cos(x + 2\pi)\): \(\cos(x)\) is periodic.
  • \(\cos(x)\) is not injective.

Then \(\cos : [0, \pi] \to \mathbb{R}\)

The function is injective and surjective in \([-1, 1]\).

So we can define \(\cos^{-1} : [-1, 1] \to [0, \pi]\)​\(\text{ such that }\cos(x)=t\) where \(t\mapsto x\)

We denote \(\arccos(t) := \cos^{-1}(t)\).

Thus \(\cos(\arccos(t)) = t \quad \forall \, t \in [-1, 1]\)

\(\arccos(\cos(x)) = x \quad \forall \, x \in [0, \pi]\)

We want to apply the same theorem:

  • \(\cos(x)\) is continuous in \((0, \pi)\).
  • \((\cos(x))'\) is continuous in \((0, \pi)\).
  • \((\cos(x))' \neq 0\) in \((0, \pi)\)​
  • \(-\sin(x) \neq 0\) in \((0, \pi)\).

Applying the theorem, \(\arccos(t)\) is differentiable and \((\arccos(t))' = \frac{1}{-\sin(\arccos(t))} = -\frac{1}{\sqrt{1 - t^2}}\)

From the unit circle:

\(\cos(s)=\frac{a}{1},\quad\sin(s)=\frac{b}{1},\quad b>0\text{ for }0<s<\pi\)

Using \(\cos^2(s) + \sin^2(s) = 1\): \(\sin^2(s) = 1 - \cos^2(s) \implies \sin^2(s) = 1 - (\cos(\arccos(t)))^2 = 1 - t^2\)

Then \(\sin(s) = \sqrt{1 - t^2}\)

Summary

\((\sin(x))' = \cos(x)\)

\((\cos(x))' = -\sin(x)\)

\((\tan(x))' = 1 + \tan^2(x) = \frac{1}{\cos^2(x)}\)

\((\arcsin(x))' = \frac{1}{\sqrt{1 - x^2}}\)

\((\arctan(x))' = \frac{1}{1 + x^2}\)

\((\arccos(x))' = -\frac{1}{\sqrt{1 - x^2}}\)