Trying to solve a related rates problem I am trying to solve a related rates problem. The problem states:
If $y = 4x -x^3$ and the x-coordinate is increasing at the rate of 1/3 unit/sec. How fast is the slope of the graph changing at the instant when $x = 2$?
I have done this:
Let $\frac {dx}{dt} = \frac {1}{3}$ unit/sec
I derive the formula:
$\frac {dy}{dt} = 4 \frac {dx}{dt} - 3x^2 \frac {dx}{dt}$ 
I substitute to solve for $dy \over dt$.
$\frac {dy}{dt} = 4(\frac {1}{3}) - 3(2^2)(\frac {1}{3}) = \frac {-8}{3}$
and then I solve for the slope as (dy/dt)/(dx/dt) = (-8/3)/3 which is -8 unit/sec
But the answer is supposed to be -4 units/sec
What am I doing wrong?
Ted
 A: Hint: How fast the slope is changing is $\frac{dw}{dt}$ where $w=\frac{dy}{dx}$.  So you will be differentiating again.
Added: The question does not ask for the slope, it asks for the rate of change of the slope. We have $\frac{dy}{dx}=4-3x^2$, so the slope is $4-3x^2$. 
For the rate of change of this, differentiate $4-3x^2$ with respect to $t$. We get that 
$$\frac{d}{dt}\left(\frac{dy}{dx}\right)=-6x\frac{dx}{dt}.$$
Since $\frac{dx}{dt}=\frac{1}{3}$, when $x=2$ we have
$$\frac{d}{dt}\left(\frac{dy}{dx}\right)=(-6)(2)(1/3)=-4.$$ 
A: A slightly different explanation that some may follow more easily:
The trick here is reading the problem closely.  It asks for the rate of change of the slope (which is itself a rate of change).  The derivative of the function will give you the slope.  The second derivative will give you the rate of change of the slope (the slope of the slope).
Since the question is asking about the change in the slope of the equation $y = 4x-x^3$, the first order of business would be to get the equation for the slope (which is the derivative, $y^{'}$ or $\frac {dy}{dx}$):
$$y = 4x-x^3$$
$$ \frac {dy}{dx} = 4 - 3x^2$$
This is the equation for the slope of the function.  So we don't let confusing notation get in the way, let's just call this equation for slope $s$.  So, $s = 4 - 3x^2$.  
Now we want to know how the slope is changing with time.  We know that $x$ is increasing at a rate of $1 \over 3$ units/sec and we want to know how the slope changes with small changes in $x$.  In other words, we were given $\frac {dx}{dt}$ and we want to find $\frac {ds}{dt}$.  So, we take the derivative of $s$ with respect to time:
$$\frac {ds}{dt} = -6x \frac {dx}{dt}$$
Substituting known values, 
$$\frac {ds}{dt} = -6 \cdot 2 \cdot \frac {1}{3} = -4$$
