Calculus I

Differentiation Rules

$\displaystyle\frac{d}{dx}\left[ c f(x) \right] = c f'(x)$
$\displaystyle\frac{d}{dx}\left[ f(x) + g(x) \right] = f'(x) + g'(x)$ and $\displaystyle\frac{d}{dx}\left[ f(x) - g(x) \right] = f'(x) - g'(x)$
$\displaystyle\frac{d}{dx}\left[ f(x) g(x) \right] = f'(x)g(x) + f(x) g'(x)$
$\displaystyle\frac{d}{dx}\left[ \frac{f(x)}{g(x)} \right] = \frac{f'(x)g(x) - f(x) g'(x)}{[g(x)]^2}$ and $\displaystyle\frac{d}{dx}\left[ \frac{1}{f(x)} \right] = - \frac{f'(x)}{[f(x)]^2}$
$\displaystyle\frac{d}{dx}\left[ f(g(x)) \right] = f'(g(x))\cdot g'(x)$ or $\displaystyle\frac{dy}{dx} = \displaystyle\frac{dy}{du} \cdot \displaystyle\frac{du}{dx}$ with and