Intuitive meaning of second, third and fourth derivatives at a point. Can someone explain me the intuitive meaning of second, third and fourth derivatives of a function say, $f(x)$ at a point (say, $a$)? I know it's hard to explain to someone novice like me! But an intuitive answer of this question can help many people.
 A: The second derivative can be thought of as the acceleration of a function,  it's the rate of change of the rate of change....positive means the function's slopes are increasing, negative means the function's slopes are decreasing.  If the function was a position function,  this would in fact be acceleration.
The third derivative can be thought of as the 'jerk' of the function.   This comes from the physical sensations of change in acceleration with respect to time:  We don't feel velocity (First derivative) at all.  We feel acceleration as a constant force.   Changes in acceleration feel like 'jerks',  that's that you feel when you are riding in a car, all the little tiny accelerations of the car's tires on the road causing faster/slower/left/right/up/downs, etc.
The fourth derivative...is probably beyond intuitive physical descriptions on this chain :)
A: To remain intuitive, the Taylor expansion of a function around a point $x_0$ is nothing else than a polynomial of grade - let's say - $n$ that has as coefficients the various derivatives of the function at that point.
Of course, the $k$-th derivative of the function and of the polynomial will coincide at $x_0$. Thus, the $k$-th derivative of a function has the meaning of the coefficient of $x^k$ of the polynomial given by the taylor expansion, which is tangent at the function in $x_0$. Also more explicitly, the equation 
$$
p_k(x) - f(x) = 0
$$
will have a root of multiplicity $k$ at $x_0$ if $p_k(x)$ is the Taylor expansion to order $k$ of $f(x)$.
For $k=2$ this corresponds to the curvature of the circle tangent to the function in $x_0$ called "osculator circle".
