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Calculus is mainly about calculating the area under a curve. We set a starting point on the X-axis and set x as the moving endpoint on the X-axis.

If the curve is f(x), and the area under the curve is A(x). We have dA = f(x)dx, where dx is a very tiny length on the X-axis, f(x)dx is the area of the very thin rectangle at x. dA means the very small change of function A from x to x+dx. From dA=f(x)dx, we have dA/dx = f(x). We call f(x) the derivative of function A.

To calculate the derivative of A: \frac {dA}{dx} as \frac{A(x+dx)-A(x)}{dx}. By some transformation we can get the result. For example, A(x) = x^2, \frac{dA}{dx} = \frac{(x+dx)^2-x^2}{dx} = \frac{2xdx + d^2x}{dx}= 2x + dx. Since dx is very small, so \frac{dA}{dx} = 2x.

Don't think dx as infinitely small, but think of it as a comparatively small value.

Derivative Rules

Three rules of derivative.

  • Constant rule: f'(x) = 0
  • Sum rule: (\alpha f(x)+ \beta g(x))' = \alpha f'(x) + \beta g'(x)
  • Product rule: (f(x)g(x))' = f'(x)g(x) + f(x)g'(x)
  • Quotient rule: \left({\frac {f(x)}{g(x)}}\right)'={\frac {f'(x)g(x)-f(x)g'(x)}{g^{2}(x)}}
  • Chain rule: (f(g(x)))' = f'(g(x))g'(x)

Mathimatical Constant

A simple definition of the mathimatical constant e is (e^x)' = e^x.

Implicit Differentiation

For example, f(x, y)=g(x, y). From this equation, we cannot directly see a form of y=.... We cannot directly calculate the derivative. What we know is this equation defines a curve (or some dots or something) in the plane.

We can conclude f'(x, y)=g'(x, y). Two questions needs to be answered.

  • Is f'(x, y) the derivative of x or y? It doesn't matter, since dy can be seen as a function of dx, and vice versa.
  • Why they still equal when calculate derivative? The curve defined by f(x, y)=g(x, y) constrain x and y to have a certain relationship. It might be the intervals, where f and g overlap. We don't need to consider single points overlaps since there is no derivative if they are scattered single points. Therefore, the equation is a definition of some curve in some intervals. As long as point (x, y) is on the curve, we have f(x, y)=g(x, y). Therefore, when we move a little bit f changes by df, which is f + df, on the right of the equation is g + dg. Since f=g and f +df = g + dg, df =dg.