On the meaning of the second derivative When we want to find the velocity of an object we use the derivative to find this. However, I just learned that when you find the acceleration of the object you find the second derivative. 
I'm confused on what is being defined as the parameters of acceleration. I always thought acceleration of an object is it's velocity (d/t). 
Furthermore, in the second derivative are we using the x value or the y value of interest. In the first derivative we were only concerned with the x value. Does this still hold true with the second derivative?
I would post pictures but apparently I'm still lacking 4 points. 
 A: The velocity is the rate of change of displacement.
The acceleration is the rate of change of velocity. 
So the velocity is the derivative with respect to $t$ of the displacement function $s(t)$. In symbols, $v(t)=s'(t)$. 
The acceleration is the derivative of velocity with respect to $t$. In symbols, $a(t)=v'(t)$.
It follows that $a(t)$ is the second derivative of displacement. In symbols, $a(t)=s''(t)$.  
If you prefer Leibniz notation, let $s$ be displacement at time $t$. Then the velocity is $\dfrac{ds}{dt}$ and the acceleration is $\dfrac{d}{dt}\left(\dfrac{ds}{dt}\right)$, which is $\dfrac{d^2s}{dt^2}$. 
A: Acceleration is very much not velocity. It's the rate at which velocity changes.
A: Acceleration is the rate of change of velocity with respect to time.  If I'm driving at a constant velocity of $30$ mph, acceleration is $0$. If I brake suddenly, my velocity drops dramatically: my velocity changes, i.e., I decelerate.  On the other hand, if I want to pass a car, or am late for work, and step on the gas pedal to accelerate, my velocity would increase from say, 30 mph to $45$ mph, or more (depending on how late I am!) Acceleration is a measure of the rate of change in velocity.
So it is $\frac{d}{dt}(v(t))$, where $v(t) = dx/dt$ is the rate of change of position with respect to time.
So we have that acceleration is the derivative of a derivative: the second derivative with respect to position, or the derivative of velocity. 
A: The derivative of the derivative is the second derivative.  Here the derivatives are with respect to time ($t$).  The dependent variable represents (one coordinate of) the position.  That might be called $x$ or $y$ or $z$, depending on what you're interested in.
A: The position function is typically denoted $r=x(t)$. Velocity is the derivative of the position function with respect to time: $v(t)=\dfrac{dx(t)}{dt}$. Acceleration is the derivative of the velocity function with respect to time: $a(t)=\dfrac{dv(t)}{dt}$. This is equivalent to the second derivative of the position function with respect to time: $$\dfrac{d}{dt}\dfrac{d}{dt}x(t)=\dfrac{d}{dt}\dfrac{dx(t)}{dt}=\dfrac{d}{dt}v(t)=\dfrac{dv(t)}{dt}=a(t).$$
The derivative is taken because it gives the change of a function with respect to its input variable.
A: Slope of a Displacement vs Time graph gives you the velocity of the body. Also, the slope of the Velocity vs Time graph gives you the acceleration of the body.
So when you find the double derivate of the displacement vs time graph, you get acceleration. When you are finding the double derivative, what you are actually doing is finding the slope of the slope of displacement vs time graph, i.e, slope of the velocity vs time graph which is nothing but acceleration.
A: Let $f(x)$ be a twice differentiable function.
Then, $f'(x)$ gives us the instantaneous rate of change of the function $f(x)$ at $x$ and $f''(x)$ gives us the instantaneous rate of change of the function $f'(x)$ at $x$.
If $f(x)$ happens to be a displacement function (say of a particle), then $f'(x)$ describes the velocity of that particle at $x$ and $f''(x)$ gives us its acceleration at that instant. Incidentally, if $f'''(x)$ exists, then it is a measure of what we call jerk.  
