Which type of number is this? 7/3

: $f(x) = x^{2} \sin x$

Step 1: Product rule $(uv)' = u'v + uv'$, where $u = x^{2}$, $v = \sin x$.

$u' = 2x$, $v' = \cos x$.

$f'(x) = 2x \sin x + x^{2} \cos x$.

f'(x) = 2x \sin x + x^{2 \cos x}

: $y = \dfrac{x}{x^{2} + 1}$

Step 1: Quotient rule $\left( \dfrac{u}{v} \right)' = \dfrac{u'v - uv'}{v^{2}}$, where $u = x$, $v = x^{2} + 1$.

$u' = 1$, $v' = 2x$.

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

Step 2: Simplify numerator: $x^{2} + 1 - 2x^{2} = 1 - x^{2}$.

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

y' = \dfrac{1 - x^{2}{(x^{2} + 1)^{2}}}

: $f(x) = e^{x} \cos x$

Step 1: Product rule $(uv)' = u'v + uv'$, where $u = e^{x}$, $v = \cos x$.

$u' = e^{x}$, $v' = -\sin x$.

$f'(x) = e^{x} \cos x + e^{x} (-\sin x) = e^{x} (\cos x - \sin x)$.

f'(x) = e^{x (\cos x - \sin x)}

: $y = \sqrt{x^{2} + 1}$

Step 1: Rewrite $y = (x^{2} + 1)^{1/2}$.

Chain rule $y' = \dfrac{1}{2} (x^{2} + 1)^{-1/2} \cdot 2x = \dfrac{x}{\sqrt{x^{2} + 1}}$.

y' = \dfrac{x{\sqrt{x^{2} + 1}}}

: $y = \sin^{3} x$

Step 1: Chain rule, let $u = \sin x$, $y = u^{3}$.

$\dfrac{dy}{dx} = 3u^{2} \dfrac{du}{dx} = 3 \sin^{2} x \cos x$.

y' = 3 \sin^{2 x \cos x}

: $y = \dfrac{\tan x}{\sec x}$

Step 1: Rewrite $y = \dfrac{\sin x / \cos x}{1 / \cos x} = \sin x$.

$y' = \cos x$.

y' = \cos x

: Tangent line to $y = x^{2}$ at $x = 1$.

Step 1: $y' = 2x$, slope at $x=1$: $m = 2(1) = 2$.

Point: $(1, 1)$.

Equation: $y - 1 = 2(x - 1)$.

$y = 2x - 1$.

y = 2x - 1

: Implicit $x^{2} + y^{2} = 25$, find $\frac{dy}{dx}$.

Step 1: Differentiate: $2x + 2y y' = 0$.

$2y y' = -2x$.

$y' = -\dfrac{x}{y}$.

\dfrac{dy{dx} = -\dfrac{x}{y}}

: Ladder 10 ft long against wall. Bottom slides away at 1 ft/s, top speed when bottom 6 ft from wall.

Step 1: Let $x$ distance from wall (bottom), $y$ height on wall (top). $x^{2} + y^{2} = 10^{2} = 100$.

Differentiate: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$.

$\frac{dy}{dt} = -\dfrac{x}{y} \frac{dx}{dt}$.

When $x=6$ ft, $y = \sqrt{100 - 36} = 8$ ft.

$\frac{dx}{dt} = 1$ ft/s.

$\frac{dy}{dt} = -\dfrac{6}{8} (1) = -0.75$ ft/s.

-0.75 ft/s