The Derivative from the Definition

$f'(x) = \lim_{h \to 0} \dfrac{\left[3(x + h)^2 - 5(x + h) + 2\right] - \left[3x^2 - 5x + 2\right]}{h}$.

$3(x + h)^2 - 5(x + h) + 2 = 3x^2 + 6xh + 3h^2 - 5x - 5h + 2$.

Numerator: $6xh + 3h^2 - 5h$.

$f'(x) = \lim_{h \to 0} \dfrac{h(6x + 3h - 5)}{h} = \lim_{h \to 0} (6x + 3h - 5)$.

$= 6x - 5$.

$g'(x) = \lim_{h \to 0} \dfrac{\left[(x + h)^3 + 4(x + h)\right] - \left[x^3 + 4x\right]}{h}$.

$(x + h)^3 + 4(x + h) = x^3 + 3x^2h + 3xh^2 + h^3 + 4x + 4h$.

Numerator: $3x^2h + 3xh^2 + h^3 + 4h$.

$g'(x) = \lim_{h \to 0} \dfrac{h(3x^2 + 3xh + h^2 + 4)}{h} = \lim_{h \to 0} (3x^2 + 3xh + h^2 + 4)$.

$= 3x^2 + 4$.

$f'(x) = \lim_{h \to 0} \dfrac{1}{h}\left[\dfrac{4}{1 - 3x - 3h} - \dfrac{4}{1 - 3x}\right]$.

$= \lim_{h \to 0} \dfrac{1}{h} \cdot \dfrac{4(1 - 3x) - 4(1 - 3x - 3h)}{(1 - 3x - 3h)(1 - 3x)}$.

Numerator: $4 - 12x - 4 + 12x + 12h = 12h$.

$f'(x) = \lim_{h \to 0} \dfrac{12}{(1 - 3x - 3h)(1 - 3x)}$.

$= \dfrac{12}{(1 - 3x)^2}$.

$f'(x) = \lim_{h \to 0} \dfrac{1}{h}\left[\dfrac{2(x + h)}{x + h + 5} - \dfrac{2x}{x + 5}\right]$.

$= \lim_{h \to 0} \dfrac{1}{h} \cdot \dfrac{2(x + h)(x + 5) - 2x(x + h + 5)}{(x + h + 5)(x + 5)}$.

Numerator: $(2x^2 + 10x + 2xh + 10h) - (2x^2 + 2xh + 10x) = 10h$.

$f'(x) = \lim_{h \to 0} \dfrac{10}{(x + h + 5)(x + 5)}$.

$= \dfrac{10}{(x + 5)^2}$.

$f'(x) = \lim_{h \to 0} \dfrac{1}{h}\left[\dfrac{1}{(x + h)^2 + 1} - \dfrac{1}{x^2 + 1}\right]$.

$= \lim_{h \to 0} \dfrac{1}{h} \cdot \dfrac{(x^2 + 1) - \left[(x + h)^2 + 1\right]}{\left[(x + h)^2 + 1\right](x^2 + 1)}$.

Numerator: $x^2 + 1 - x^2 - 2xh - h^2 - 1 = -h(2x + h)$.

$f'(x) = \lim_{h \to 0} \dfrac{-(2x + h)}{\left[(x + h)^2 + 1\right](x^2 + 1)}$.

$= \dfrac{-2x}{(x^2 + 1)^2}$.

$f'(x) = \lim_{h \to 0} \dfrac{\sqrt{2x + 2h + 1} - \sqrt{2x + 1}}{h}$.

$= \lim_{h \to 0} \dfrac{\sqrt{2x + 2h + 1} - \sqrt{2x + 1}}{h} \cdot \dfrac{\sqrt{2x + 2h + 1} + \sqrt{2x + 1}}{\sqrt{2x + 2h + 1} + \sqrt{2x + 1}}$.

Numerator: $(2x + 2h + 1) - (2x + 1) = 2h$.

$f'(x) = \lim_{h \to 0} \dfrac{2}{\sqrt{2x + 2h + 1} + \sqrt{2x + 1}}$.

$= \dfrac{2}{2\sqrt{2x + 1}} = \dfrac{1}{\sqrt{2x + 1}}$.

$g'(x) = \lim_{h \to 0} \dfrac{\sqrt{5 - x - h} - \sqrt{5 - x}}{h}$.

$= \lim_{h \to 0} \dfrac{\sqrt{5 - x - h} - \sqrt{5 - x}}{h} \cdot \dfrac{\sqrt{5 - x - h} + \sqrt{5 - x}}{\sqrt{5 - x - h} + \sqrt{5 - x}}$.

Numerator: $(5 - x - h) - (5 - x) = -h$.

$g'(x) = \lim_{h \to 0} \dfrac{-1}{\sqrt{5 - x - h} + \sqrt{5 - x}}$.

$= -\dfrac{1}{2\sqrt{5 - x}}$.

$f'(x) = \lim_{h \to 0} \dfrac{1}{h}\left[\dfrac{1}{\sqrt{x + h}} - \dfrac{1}{\sqrt{x}}\right]$.

Combine first: $\lim_{h \to 0} \dfrac{\sqrt{x} - \sqrt{x + h}}{h\sqrt{x}\sqrt{x + h}}$. Then multiply top and bottom by $\sqrt{x} + \sqrt{x + h}$.

Numerator: $x - (x + h) = -h$.

$f'(x) = \lim_{h \to 0} \dfrac{-1}{\sqrt{x}\sqrt{x + h}\left(\sqrt{x} + \sqrt{x + h}\right)}$.

$= \dfrac{-1}{\sqrt{x} \cdot \sqrt{x} \cdot 2\sqrt{x}} = -\dfrac{1}{2x^{3/2}}$.

$f'(5) = \lim_{h \to 0} \dfrac{f(5 + h) - f(5)}{h}$.

$f(5 + h) = 25 + 10h + h^2 - 15 - 3h = h^2 + 7h + 10$ and $f(5) = 25 - 15 = 10$.

$f(5 + h) - f(5) = h^2 + 7h = h(h + 7)$.

$f'(5) = \lim_{h \to 0} (h + 7) = 7$.

$f'(9) = \lim_{h \to 0} \dfrac{\sqrt{9 + h} - \sqrt{9}}{h}$.

$f(9 + h) = \sqrt{9 + h}$ and $f(9) = 3$.

Conjugate: $\dfrac{\sqrt{9 + h} - 3}{h} \cdot \dfrac{\sqrt{9 + h} + 3}{\sqrt{9 + h} + 3} = \dfrac{(9 + h) - 9}{h\left(\sqrt{9 + h} + 3\right)} = \dfrac{h}{h\left(\sqrt{9 + h} + 3\right)}$.

$f'(9) = \lim_{h \to 0} \dfrac{1}{\sqrt{9 + h} + 3} = \dfrac{1}{6}$.

$f'(2) = \lim_{x \to 2} \dfrac{x^3 - f(2)}{x - 2}$.

$f(2) = 8$, so the quotient is $\dfrac{x^3 - 8}{x - 2}$.

Difference of cubes: $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$.

$f'(2) = \lim_{x \to 2} (x^2 + 2x + 4) = 4 + 4 + 4 = 12$.

It has the shape $\dfrac{f(a + h) - f(a)}{h}$.

$f(x) = x^4$, $a = 3$; check $f(3) = 81$ ✓.

$f'(x) = 4x^3$.

$f'(3) = 4 \cdot 27 = 108$.

It has the shape $\dfrac{f(x) - f(a)}{x - a}$.

$f(x) = \sin x$, $a = \tfrac{\pi}{3}$; check $\sin\tfrac{\pi}{3} = \tfrac{\sqrt{3}}{2}$ ✓.

$f'(x) = \cos x$.

$f'\!\left(\tfrac{\pi}{3}\right) = \cos\tfrac{\pi}{3} = \tfrac{1}{2}$.

It has the shape $\dfrac{f(a + h) - f(a)}{h}$.

$f(x) = \log x$, $a = e$; check $\log e = 1$ ✓.

$f'(x) = \dfrac{1}{x}$.

$f'(e) = \dfrac{1}{e}$.

Left value $\sqrt{4} = 2$, right value $4a + b$: $4a + b = 2$.

Left: $\left(\sqrt{x}\right)' = \dfrac{1}{2\sqrt{x}}$. Right: $(ax + b)' = a$.

Left slope $\dfrac{1}{2\sqrt{4}} = \dfrac{1}{4}$. Right slope $a$.

$a = \dfrac{1}{4}$.

$b = 2 - 4 \cdot \tfrac{1}{4} = 1$.

Both sides give $|4 - 4| = 0 = g(2)$, so $g$ is continuous at $2$.

Just left of $2$, $x^2 - 2x = x(x - 2) < 0$, so $g = 2x - x^2$ and $g' = 2 - 2x$. Right of $2$, $g = x^2 - 2x$ and $g' = 2x - 2$.

Left slope $2 - 4 = -2$, right slope $4 - 2 = 2$.

$-2 \neq 2$: the graph has a corner at $x = 2$.

$g$ is continuous but not differentiable at $x = 2$.

$k(x) = -x^3$ for $x < 0$ and $x^3$ for $x \geq 0$; both give $0 = k(0)$, so $k$ is continuous at $0$.

Left: $(-x^3)' = -3x^2$. Right: $(x^3)' = 3x^2$.

Both slopes are $0$ at $x = 0$. By the definition: $\lim_{h \to 0} \dfrac{h^2|h| - 0}{h} = \lim_{h \to 0} h|h| = 0$.

$0 = 0$: the slopes match, no corner.

$k$ is differentiable at $0$, with $k'(0) = 0$.

Left value $1^2 = 1$, right value $f(1) = 2$: not continuous at $x = 1$. Stop here.

Not needed. (The derivatives would be $2x$ and $2$.)

Not needed — and the slopes are both $2$, which proves nothing without continuity.

Nothing to compare: a jump already rules out a tangent line.

$f$ is not differentiable at $x = 1$, because it is not continuous there.