Calculus

The derivative of a function ƒ is the function ƒ' whose values are the slope of the tangent line at each point on the curve of ƒ.
The integral of a function ƒ is the function F whose values' difference is the area under the curve of ƒ in an interval.

Would like to know at each point on the curve of function ƒ what is the rate of change, i.e. by how much the function is increasing or decreasing at that point instantaneously. This would be the slope of the line tangent to (i.e. just touching at) the curve at that point. But on a curve, unlike on a line, the slope is different at each point.
Between the endpoints of an interval around a point the average rate of change can be calculated: a secant line connects the endpoints; its slope Δy/Δx is the rate of change between the endpoints, ie. the average rate of change; Shrinking the interval's width to[ward] zero makes the secant line become the tangent line to/at the point. As Δx→0 it becomes dx, the differential of x, ("d for darn small"); Δy becomes dy and dy/dx is the slope of the tangent line. The dx is infinitesimally small ("a ghost of a departed quantity"); or it means that the denominator approaches 0. The tangent line (that touches the curve at that point) is the rate of change of the function at that point. Its slope is the change of y per change of x, i.e. the rate of change. This is the instantaneous rate of change, i.e. the rate of change at that single point.
The derivative of a function ƒ is the function ƒ' whose values are the slope of the tangent line at each point on the curve of ƒ. I.e. from your ƒ function you derive another function ƒ' which tells you the slopes of the tangent lines at every point on the curve of ƒ. Given an x, ƒ(x) is the y value of a point on the curve of ƒ. ƒ'(x) is the slope m of the line tangent to that (x,y) point on the curve of ƒ.
The tangent line (i.e. its slope) is defined by the derivative. The tangent line is the best approximation of the curve. This slope is the best constant approximation for a rate of change around the point. This slope is what is meant by the instantaneous rate of change.
Differentiation (d/dx) is the process of determing what the derivative function is, i.e. taking the derivative.
A function must be "nice" to have a derivative (at a point): it must be continuous (i.e. no holes, jumps, vertical asymptotes) and differentiable (i.e. no kinks, cusps).

Various rules and formulas exist for finding the derivative of categories of function, e.g. polynomial, trigonometric, exponential etc.
(cf)' = cf'
Power rule: (xr)' = rxr-1
(f±g)' = f'±g'
Product rule: (fg)'= f'g + fg'     (fgh)'= f'gh + fg'h + fgh' ...
Quotient rule: (f/g)' = (f'g-fg')/g2     (1/f)'=-f'/f2
Chain rule: f(g(x))' = f'(g(x))g'
sin' = cos     cos'=-sin     tan'=sec2
(ex)' = ex     (bx)' = (ln b)bx
ln' = 1/x     logb' = 1/(ln b ln x)

From algebra, know that x-intercepts of function ƒ are the solutions to ƒ(x)=0. y-intercept = ƒ(0).
Derivative indicates where ƒ is increasing, decreasing, or horizontal:
ƒ'(x)>0 ie. derivative/slope is positive, means ƒ is increasing at (x,ƒ(x)).
ƒ'(x)<0 ie. derivative/slope is negative, means ƒ is decreasing at (x,ƒ(x)).

ƒ'(x)=0 ie. derivative/slope is zero (ƒ' x-intercept), means ƒ is horizontal at (x,ƒ(x)). Solve ƒ'(x)=0 to find x-intercepts of ƒ'.
X's where the derivative is zero are critical numbers c which might be extrema, i.e. (local) maximum or minimum values of the function.

The second derivative ƒ''(x), the derivative of the derivative.
indicates the concavity/"bending" of the function ƒ:
ƒ''(x)>0 ie. 2nd derivative is positive, means ƒ is concave up, tangent line is below ƒ.
ƒ''(x)<0 ie. 2nd derivative is negative, means ƒ is concave down, tangent line is above ƒ.

ƒ''(x)=0 ie. 2nd derivative is zero (ƒ'' x-intercept), [might be] inflection point where concavity changes. Solve ƒ''(x)=0 to find x-intercepts of ƒ''.
At a critical number c, if ƒ''(c)>0 the extremum is a minimum.
At a critical number c, if ƒ''(c)<0 the extremum is a maximum.

Another f function example:



Examples of first and second derivatives:

Would also like to know the area under the curve of the function ƒ over any interval [a,b]. Integration ∫ of the function to obtain its integral F.
F(b)-F(a) = area under the curve of function ƒ in the interval [a,b].
This area could represent something physical depending on the application, such as time, energy, money, distance etc.

Fundamental theorem of calulus: differentiation and integration are related:

Examples