The tangent of an angle is it's gradient or slope: $$ \tan{\theta} = \frac{oppostite}{adjacent} = \frac{y}{x} $$

Loosely speaking a tangent line just touches a curve and the gradient of the tangent line is the derivative of the curve at that point. So the derivative (\(f'(x)\)) gives the changing slope of the a curve. The integral of the curve, \(\int f(x) dx\), on the other hand, gives the area under the curve.

Gradient: $$m = \frac{\Delta y}{\Delta x} $$ $$m = \frac{\Delta f(x)}{\Delta x} = \frac{f(x+h) - f(x)}{(x+h) - x} = \frac{f(x+h) - f(x)}{h} $$

The derivative: $$ \frac{dy}{dx} = f'(x) = \lim_{h\to a} \frac{f(x+h) - f(x)}{h} $$

There are various notations for a derivative: $$ f'(x) = \frac{dy}{dx} = y', y=f(x) $$

For parametric equations, where 'x' and 'y' depend on another variable 't', for instance the position of a vehicle at time 't', dot notation can be used: $$ \dot{x} = \frac{x}{t}, \dot{y} = \frac{y}{t} $$

The fundemental theorem of calculus relates the derivative and the integral: $$\int^x_{-\infty} f'(t) dt = f(x)$$

Vector

If a function has more than one paramater, e.g. f(x,y) then we can find the derivative by 'x' or by 'y', these are called partial derivatives: $$ \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} $$

The 'del' or 'nabla' notation gives the derivative of a vector as a vector of partial derivatives: $$ \nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}) $$

The gradient (grad), divergence (div) and curl operators are defined using 'nabla': $$ grad f = \nabla f $$ $$ div \vec{v} = \nabla . \vec{v} $$ $$ curl \vec{v} = \nabla \times \vec{v} $$

Grad is the verctor derivative of a scalar field, pointing in the direction of greatest increase of 'f' with a magnitude equal to the rate of increase (like a derivative). Given a height function $$h(x,y)$$ then 'grad' gives the steepest slope.

Div gives a scalar ...

Curls gives a vector ...

The laplacian operator gives a scalar: $$\Delta = \nabla . \nabla = \nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} $$

...

Given the unit circle \(|z| = 1\) as the contour, let \(z(t)=e^{it}\) where \(t\in[0,2\pi]\), the contour integral: $$\oint_C = \frac{1}{z}dz = \int_0^{2\pi} \frac{1}{e^{it}}ie^{it} dt = i \int_0^{2\pi} 1dt = [t]_0^{2\pi}i = (2\pi - 0)i = 2\pi i$$

...

Limits

Given assumptions L'Hopital's rule: $$\lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)}$$

...

The rule for differentiating the product of two factors is: $$(uv)' = u'v + v'$$ This is intimately related to the rule for integration by parts: $$\int_a^b u(x) v(x)' dx = [u(x)v(x)]^b_a - \int_a^b u(x)'v(x) dx$$ Or more compactly: $$\int uv' dx = uv - \int_a^b u'v dx$$ Rearrange: $$uv = \int uv' dx + \int_a^b u'v dx$$ Differentiate both sides and we obtain the product rule. $$(uv)' = uv' + u'v$$ The fundemental theorem of calculus: $$\int f'(x) dx = f(x)$$

The chain rule for differentiating the function of a function is: $$\frac{d u(v(x))}{dx} = \frac{du}{dv} \frac{dv}{dx}$$ This is intimately related to the rule for integration by substitution, sometimes called the reverse chain rule: $$\int f(g(x)g'(x) dx = \int f(u) du$$

...

Newton's method is used to find the roots for an equation, where y(x) =0

Start with an approximation for the derivative: $$y'(x) = \frac{dy}{dx} \approx \frac{\Delta y}{\Delta x} = \frac{y(x_1)-y(x_0)}{x_1-x_0}$$

Since we are trying to fined the roots of the equation, where y(x) = 0, then let y(x_1) = 0:

$$y'(x) \approx \frac{0-y(x_0)}{x_1-x_0}$$ $$x_1-x_0 \approx \frac{-y(x_0)}{y'(x)}$$ $$x_1 = x_0 - \frac{y(x_0)}{y'(x)}$$

Guess a value for x₀ (the initial value) and then the following should produce successively better approximations: $$x_{n+1} = x_{n} - \frac{y(x_n)}{y'(x)}$$