Tangent Line and derivative

Let $f (x)$ be defined on open interval $ℐ$ that contains $ω$ . Let $ℎ \neq 0$ such that $x = ω + ℎ \in ℐ$ , then

𝒟 = \frac{f (x) - f (ω)}{x - ω} = \frac{f (ω + ℎ) - f (ω)}{ℎ}

is the difference quotient with increment $ℎ$ .

Let $f (x)$ be defined on open interval $ℐ$ that contains $ω$ . The tangent line to $f (x)$ at the point $𝒫 = (ω, f (ω))$ is the line that passes through $𝒫$ and has slope:

𝓂 = \lim_{x \to ω} \frac{f (x) - f (ω)}{x - ω} = \lim_{ℎ \to 0} \frac{f (ω + ℎ) - f (ω)}{ℎ}

Derivative at a point

The derivative of $f (x)$ at $ω$ , denoted by $f' (ω)$ is

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

Derivative as a function

Let $f (x)$ be a function. The derivative of $f (x)$ with respect to $x$ , denoted by $f' (x)$ is the function

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

defined for all values $x$ for which the limit exists.

f' (x) = \lim_{𝓏 \to x} \frac{f (𝓏) - f (x)}{𝓏 - x}

Notation

$y = f (x)$

$y' = f' (x) = \frac{d y}{d x} = \frac{d f}{d x} = \frac{d}{d x} f (x)$
$f' (ω) = \frac{d y}{d x} |_{x = ω} = \frac{d f}{d x} |_{x = ω} = \frac{d}{d x} f (x) |_{x = ω}$

$f (x)$ is differentiable at $ω$: $f' (ω)$ exists
$f (x)$ is differentiable on $ℐ$: $\forall ω \in ℐ, f' (ω)$ exists
$f (x)$ is differentiable: $f' (ω)$ exists for all $ω$ in domain of $f (x)$

right hand derivative at $ω$: $\lim_{ℎ \to 0^{+}} \frac{f (ω + ℎ) - f (ω)}{ℎ}$
left hand derivative at $ω$: $\lim_{ℎ \to 0^{-}} \frac{f (ω + ℎ) - f (ω)}{ℎ}$

If $f (x)$ is differentiable at $ω$ then $f (x)$ is continuous at $ω$ .

Derivative Examples

Find tangent line to $f (x) = \frac{x^{2}}{2}$ at $P = [1, \frac{1}{2}]$ .

Find tangent line to $f (x) = x^{2}$ at $x = 3$ using

$\lim_{x \to 3} \frac{f (x) - f (3)}{x - 3}$
$\lim_{ℎ \to 0} \frac{f (3 + ℎ) - f (3)}{ℎ}$

Let $f (2) = 3 x^{2} - 4 x + 1$ , evaluate

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

Find $f^{'} (0)$ for

$f (x) = x^{\frac{1}{3}}$
$f (x) = x^{\frac{2}{3}}$

Compare with $f (x) = | x |$ and $f (x) = x$ at zero.

Let $s (x) = \sqrt{-2 x^{2} + 729} = \sqrt{-2 x^{2} + 27^{2}}$ evaluate

$s^{'} (12) = \lim_{ℎ \to 0} \frac{s (12 + ℎ) - s (12)}{ℎ}$
$s^{'} (ω)$

Find $f^{'} (x)$ for

$f_{1} (x) = 5$
$f_{2} (x) = x - 4$
$f_{3} (x) = x^{2} - 3 x$
$f_{4} (x) = \sqrt{x}$

Compare the graphs of $f (x)$ and $f^{'} (x)$

If possible evaluate $f^{'} (0)$

for $f_{1} (x) = {\begin{matrix} x sin (\frac{1}{x}) & x \neq 0 \\ 0 & x = 0 \end{matrix}$
for $f_{2} (x) = {\begin{matrix} x^{2} sin (\frac{1}{x}) & x \neq 0 \\ 0 & x = 0 \end{matrix}$

Find values $b$ and $c$ such that

f (x) = {\begin{matrix} \frac{1}{10} x^{2} + b x + c & x < 10 \\ \frac{-1}{4} x + \frac{5}{2} & 10 \leq x \end{matrix}

is continuous and differentiable.

Derivative Rules

If $f (x) = c$ , then $f' (x) = 0$

\frac{d}{d x} c = 0

If $f (x) = x^{n}$ then $f' (x) = n x^{n - 1}$ .

\frac{d}{d x} x^{n} = n x^{n - 1}

[𝒶 f (x) + 𝒷 g (x)]' = 𝒶 f' (x) + 𝒷 g' (x)

\frac{d}{d x} [𝒶 f (x) + 𝒷 g (x)] = 𝒶 \frac{d}{d x} f (x) + 𝒷 \frac{d}{d x} g (x)

[f (x) g (x)]' = f' (x) g (x) + f (x) g' (x)

\frac{d}{d x} [f (x) g (x)] = [\frac{d}{d x} f (x)] g (x) + f (x) [\frac{d}{d x} g (x)]

{[\frac{f (x)}{g (x)}]}^{'} = \frac{f^{'} (x) g (x) - f (x) g^{'} (x)}{{(g (x))}^{2}}

\frac{d}{d x} [\frac{f (x)}{g (x)}] = \frac{[\frac{d}{d x} f (x)] g (x) - f (x) [\frac{d}{d x} g (x)]}{{(g (x))}^{2}}

Higher order derivatives

\begin{matrix} y = f (x) \\ {[y]}^{'} & = & y^{'} & {[f (x)]}^{'} & = & f^{'} (x) \\ {[y^{'}]}^{'} & = & y^{″} & {[f^{'} (x)]}^{'} & = & f^{″} (x) \\ {[y^{″}]}^{'} & = & y^{‴} & {[f^{″} (x)]}^{'} & = & f^{‴} (x) \\ {[y^{‴}]}^{'} & = & y^{(4)} & {[f^{‴} (x)]}^{'} & = & f^{(4)} (x) \\ ⋮ \\ {[y^{(n)}]}^{'} & = & y^{(n + 1)} & {[f^{(n)} (x)]}^{'} & = & f^{(n + 1)} (x) \end{matrix}

In Leibnitz notation

\begin{matrix} f^{″} (x) = {[f^{'} (x)]}^{'} & = \frac{d \frac{d y}{d x}}{d x} & = \frac{d^{2} y}{{d x}^{2}} \\ f^{(n)} (x) & = \frac{d^{n} y}{{d x}^{n}} \end{matrix}

Derivative Rules Examples

Find derivatives of

$f_{1} (x) = 3 x^{4}$
$f_{2} (x) = 2 x^{5} - 6 x^{2}$

Find their tangent lines at $x = -1$ .

Find k′ (2) if
- $k (x) = f (x) g (x)$
- $f (2) = 3$ , $f^{'} (2) = -4$ ,
- $g (2) = 8$ , $g^{'} (2) = 5$
Find $k^{'} (x)$ if $k (x) = (x^{2} - x) (4 x^{3} + 5 x^{2})$

geometry of product rule
Find $q^{'} (x)$ if $q (x) = \frac{5 x^{2}}{4 x - 3}$

Compute derivatives of

$k (x) = 3 f (x) + x^{2} g (x)$
$k (x) = f (x) g (x) h (x)$
$k (x) = \frac{2 x^{3} f (x)}{3 x - 3}$