Which type of number is this? 7/3

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

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

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

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

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

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

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

$x^{2} + 1 - 2x^{2} = 1 - x^{2}$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$y' = \cos x$.

$$y' = \cos x$$

y' = \cos x

: Tangent line to $y = x^{2}$ 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 $\frac{dy}{dx}$.

$2x + 2y y' = 0$.

$2y y' = -2x$.

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

$$\frac{dy}{dx} = -\frac{x}{y}$$

\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.

$x^{2} + y^{2} = 10^{2} = 100$.

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