Derivatives as functions and higher derivatives
Read a whole function of slopes, then differentiate the rate itself.
A function of slopes
The number \(f^{\prime}(a)\) is the tangent slope at one input. Letting the input vary gives the derivative function:
\[f^{\prime}(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h.\]Its domain contains exactly the inputs where this limit exists as a finite number. A point on the graph of \(f^{\prime}\) records a slope of \(f\), rather than a height of \(f\).
For example, \(f(x)=x^2\) gives \(f^{\prime}(x)=2x\): negative inputs have negative slopes, zero has a horizontal tangent, and positive inputs have positive slopes. The graph of the derivative is a line even though the original graph is a parabola.
Differentiating a rate
Definition
Higher derivatives
The second derivative is the derivative of the first derivative:
\[f^{\prime\prime}(x)=\lim_{h\to0}\frac{f^{\prime}(x+h)-f^{\prime}(x)}h.\]It measures how the slope of \(f\) changes. Repeating the process gives the third derivative \(f^{(3)}\) and, more generally, the nth derivative \(f^{(n)}\), wherever each successive derivative exists.
For \(y=f(x)\), the notations \(y^{\prime\prime}\) and \(d^2y/dx^2\) also mean the second derivative. Here \(d^2y/dx^2=d(dy/dx)/dx\); it is not the square of \(dy/dx\).
The height of the derivative graph tells you the slope of the original graph. Its slope tells you the second derivative.
If \(f^{\prime\prime}>0\) on an interval, \(f^{\prime}\) is increasing there; if \(f^{\prime\prime}<0\), it is decreasing. The sign of \(f^{\prime\prime}\) alone does not tell you whether \(f\) is rising or falling.
Position, velocity, acceleration
For motion along a straight track, position \(s(t)\) records location relative to an origin. Velocity \(v(t)=s^{\prime}(t)\) is its rate of change, and acceleration \(a(t)=v^{\prime}(t)=s^{\prime\prime}(t)\) is the rate of change of velocity. Speed is \(\lvert v(t)\rvert\): it has no direction.
Consider a cart with position, in metres,
\[s(t)=t^3-6t^2+9t,\qquad 0\le t\le4.\]The rules already introduced give
\[v(t)=3t^2-12t+9,\qquad a(t)=6t-12.\]At \(t=1.5\), both velocity and acceleration are negative: the cart moves in the negative direction and speeds up. At \(t=2.5\), velocity is still negative but acceleration is positive: it moves in the same direction while slowing down. Acceleration describes the change in velocity, not the direction of motion.
Worked examples
Example 1
Differentiate the function, then evaluate
Let \(f(x)=x^4-2x^3+x\). Successive differentiation gives
\[\begin{aligned} f^{\prime}(x)&=4x^3-6x^2+1,\\ f^{\prime\prime}(x)&=12x^2-12x,\\ f^{(3)}(x)&=24x-12,\\ f^{(4)}(x)&=24,\qquad f^{(5)}(x)=0. \end{aligned}\]At \(x=1\), \(f^{\prime}(1)=-1\) and \(f^{\prime\prime}(1)=0\). The original graph has slope \(-1\) there; the derivative graph has a horizontal tangent. These are different statements.
Example 2
Use the chain rule again
For \(y=(1+x^2)^3\), first differentiate the composition:
\[y^{\prime}=6x(1+x^2)^2.\]Now differentiate this new function, using the product and chain rules:
\[\begin{aligned} y^{\prime\prime}&=6(1+x^2)^2+6x\bigl[2(1+x^2)\cdot2x\bigr]\\ &=6(1+x^2)(1+5x^2). \end{aligned}\]The second derivative is positive for every real input, so the first derivative is increasing—even where the first derivative itself is negative.
Example 3
Keep track of domains
For \(r(x)=\sqrt{x}\), the function is defined for \(x\ge0\), but
\[r^{\prime}(x)=\frac12x^{-1/2},\qquad r^{\prime\prime}(x)=-\frac14x^{-3/2},\qquad x>0.\]There is no finite first derivative at zero, so there cannot be a second derivative there. For \(x>0\), \(r^{\prime}>0\) while \(r^{\prime\prime}<0\): the graph rises, but its slopes become less steep.