We are now accustomed to investigating the behavior of a function by examining its derivative.
Because \(f'\) is itself a function, it is perfectly feasible for us to consider the derivative of the derivative, which is the new function \(y = [f'(x)]'\text{.}\) We call this resulting function the second derivative of \(f\text{,}\) and denote the second derivative by \(y = f''(x)\text{.}\) Consequently, we will sometimes call \(f'\) the first derivative
of \(f\text{,}\) rather than simply the derivative
of \(f\text{.}\)
Second Derivative
The second derivative is the derivative of the first derivative. That is,
\begin{equation*}
f''(x) = \frac{d}{dx}\left[f'(x)\right]\text{.}
\end{equation*}
We read \(f''(x)\)
as \(f\)double prime of \(x\)
, or as the second derivative of \(f\)
.
The meaning of the derivative function still holds, so when we compute \(f''(x)\text{,}\) this new function measures slopes of tangent lines to the curve \(y = f'(x)\text{,}\) as well as the instantaneous rate of change of \(y = f'(x)\text{.}\) In other words, just as the first derivative measures the rate at which the original function changes, the second derivative measures the rate at which the first derivative changes. The second derivative will help us understand how the rate of change of the original function is itself changing.
Example2.69
Consider the function \(f(x)=x^53x^4+x\text{.}\) The derivative is:
\begin{equation*}
f'(x)=5x^412x^3+1.
\end{equation*}
This creates a new function whose derivative can also be found:
\begin{equation*}
f''(x)=\frac{d}{dx}\left[5x^412x^3+1\right]=20x^336x^2.
\end{equation*}
Example2.70
Consider the function \(g(x)=3(x^3+4x+2)^7\text{.}\) To find the derivative we must use the chain rule. The derivative is:
\begin{equation*}
g'(x)=21(x^3+4x+2)^6(3x^2+4).
\end{equation*}
To take the second derivative, first we must use the product rule:
\begin{equation*}
g''(x)=\frac{d}{dx}\left[21(x^3+4x+2)^6(3x^2+4)\right]=\frac{d}{dx}\left[21(x^3+4x+2)^6\right](3x^2+4)+21(x^3+4x+2)^6\frac{d}{dx}\left[(3x^2+4)\right]
\end{equation*}
The derivative of the first term requires using the chain rule again:
\begin{equation*}
g''(x)=126(x^3+4x+3)^5(3x^2+4)(3x^2+4)+21(x^3+4x+3)^6(6x).
\end{equation*}
Note that the term \((3x^2+4)\) appears twice, once for each derivative, both are the derivative of the inside function.
Example2.71
Find the second derivative for each each of the following functions.
\(\displaystyle p(x) = 4x^2 + 3\sqrt{x}+\frac{4}{x^2}\)
\(\displaystyle m(x) = \frac{4x}{5x+2}\)
\(\displaystyle h(x) = \sqrt{3x^2+5x}\)
\(\displaystyle c(x) = \frac{3}{(7x^3+5x)^5}\)
Answer
\(p''(x) = 8\frac{3}{3x^{3/2}}+\frac{24}{x^4}\text{.}\)
\(m''(x) = \frac{80}{(5x+2)^3} \text{.}\)
\(\displaystyle h''(x) =\frac{1}{4}(3x^2+5x)^{3/2}(6x+5)(6x+5)+\frac{1}{2}(3x^2+5x)^{1/2}(6)\text{.}\)
\(c''(x)=90(7x^3+5x)^{7}(21x^2+5)(21x^2+5)+15(7x^3+5x)^{6}(42x)\text{.}\)
Solution

First rewrite this so that the power rule applies to each term:
\begin{equation*}
p(x) = 4x^2 + 3\sqrt{x}+\frac{4}{x^2}=4x^2+3x^{1/2}+4x^{2}\text{.}
\end{equation*}
Then the first derivative is
\begin{equation*}
p'(x) = 8x + \frac32 x^{1/2}8x^{3}\text{.}
\end{equation*}
Then take the derivative again to get the second derivative:
\begin{equation*}
p''(x) = 8  \frac34 x^{3/2}+24x^{4}=8\frac{3}{3x^{3/2}}+\frac{24}{x^4}\text{.}
\end{equation*}

To get the derivative apply the quotient rule:
\begin{equation*}
m'(x)=\frac{4(5x+2)(4x)(5)}{(5x+2)^2}=\frac{20x+820x}{(5x+2)^2}=\frac{8}{(5x+2)^2}\text{.}
\end{equation*}
Simplify the numerator in order to make finding the second derivative easier. Then we can rewrite as
\begin{equation*}
m'(x)=8(5x+2)^{2}
\end{equation*}
and use the chain rule to find the second derivative.
\begin{equation*}
m''(x)=16(5x+2)^{3}(5)=\frac{80}{(5x+2)^3}\text{.}
\end{equation*}

By the chain rule,
\begin{equation*}
h'(x) = \frac{1}{2}(3x^2+5x)^{1/2}(6x+5)\text{.}
\end{equation*}
Applying both the produce rule and chain rule can find the second derivative
\begin{equation*}
h''(x) =\frac{1}{4}(3x^2+5x)^{3/2}(6x+5)(6x+5)+\frac{1}{2}(3x^2+5x)^{1/2}(6)\text{.}
\end{equation*}

First rewrite
\begin{equation*}
c(x)=\frac{3}{(7x^3+5x)^5}=3(7x^3+5x)^{5}\text{.}
\end{equation*}
Use the chain rule to find the derivative:
\begin{equation*}
c'(x)=15(7x^3+5x)^{6}(21x^2+5)
\end{equation*}
Applying both the produce rule and chain rule can find the second derivative
\begin{equation*}
c''(x)=90(7x^3+5x)^{7}(21x^2+5)(21x^2+5)+15(7x^3+5x)^{6}(42x)\text{.}
\end{equation*}
SubsectionAcceleration
If the function \(s(t)\) gives the position of an object at time \(t\) then \(s'(t)\) gives the change in position, otherwise known as velocity. That is, \(s'(t)=v(t)\text{,}\) where \(v(t)\) is the velocity function. Using the alternate notation, we have \(\displaystyle \frac{ds}{dt}=v(t)\text{.}\)
Following this same idea, \(v'(t)\) gives the change in velocity, more commonly called acceleration. Using derivative notation, \(v'(t)=a(t)\text{.}\) Therefore, \(s''(t)=a(t)\text{.}\) That is, the second derivative of the position function gives acceleration. Using the alternative notation, we write \(\displaystyle \frac{d^2s}{dt^2}=a(t)\text{.}\)
Example2.72
Suppose that an object is launched from a height of 12 feet with an initial velocity of 15 feet per second. We can describe the position of the object with the following function:
\begin{equation*}
s(t)=16^2+15t+12.
\end{equation*}
 Find an equation for the velocity of the object; make sure to include units.
 Find an equation for the acceleration of the object; make sure to include units.
Answer
 \(v(t)=32t+15 \) feet per second.
 \(a(t)=32\) feet per second per second (or \(ft/s^2\)).
Solution
 The velocity is the derivative of position:
\begin{equation*}
v(t)=s'(t)=32t+15.
\end{equation*}
The units of velocity are feet per second.
 The acceleration is the derivative of velocity or the second derivative of position:
\begin{equation*}
a(t)=v'(t)=s''(t)=32.
\end{equation*}
The units are feet per second per second, often written as \(ft/s^2\text{.}\) Here the acceleration due to gravity is constant, typically in physics acceleration due to gravity is taken as \(32ft/s^2\text{.}\) This is negative since gravity pulls an object down. If we are working in the metric system then gravity is given as \(9.8m/s^2\text{.}\)
SubsectionHigher Order Derivatives
We have seen that when we take the derivative of a function a new function \(f'(x)\) is created. We took the derivative of this new function to get the second derivative. We can repeat this process as many times as we wish.
The third derivative of a function \(f(x)\) is the derivative of the second derivative, written as \(f'''(x)\) or more commonly as \(\displaystyle \frac{d^3f}{dx^3}\text{.}\)
The fourth derivative of a function \(f(x)\) is the derivative of the third derivative, written as \(f^{(4)}(x)\) or more commonly as \(\displaystyle \frac{d^4f}{dx^4}\text{.}\)
We can continue to take derivatives in this way.
\(n^{th}\) Derivative
The \(n^{th}\) derivative is
\begin{equation*}
f^{(n)}(x) =\frac{d^nf}{dx^n}= \frac{d^n}{dx^n}\left[f(x)\right]\text{.}
\end{equation*}
This is the derivative of \(f(x)\) taken \(n\) times. Alternatively we refer to this as the \(n^{th}\) order derivative.
Example2.73
Consider the function \(f(x)=x^53x^4+x\text{.}\) Find the \(5^{th}\) derivative.
To find the \(5^{th}\) derivative we will take the derivative \(5\) times.
The first derivative is:
\begin{equation*}
f'(x)=\frac{df}{dx}=5x^412x^3+1.
\end{equation*}
The second derivative is:
\begin{equation*}
f''(x)=\frac{d^2f}{dx^2}=20x^336x^2.
\end{equation*}
The third derivative is:
\begin{equation*}
f'''(x)=\frac{d^3f}{dx^3}=60x^272x.
\end{equation*}
The fourth derivative is:
\begin{equation*}
f^{(4)}(x)=\frac{d^4f}{dx^4}=120x72.
\end{equation*}
The fifth derivative is:
\begin{equation*}
f^{(5)}(x)=\frac{d^5f}{dx^5}=120.
\end{equation*}
Two notes from Example2.73:
1) In Example2.73 any further derivatives would end as 0 since the derivative of a constant is 0. This is true of any polynomial, eventually if you take enough derivatives they will end as 0.
2) For a polynomial as you take more derivatives the function becomes simpler.
Example2.74
Consider the function \(\displaystyle g(x)=\frac{1}{x}\text{,}\) find the \(4^{th}\) derivative of \(g(x)\text{.}\)
Start with the first derivative of \(g(x)\text{:}\)
\begin{equation*}
g'(x)=x^{2}=\frac{1}{x^2}.
\end{equation*}
The second derivative is:
\begin{equation*}
g''(x)=2x^{3}=\frac{2}{x^3}
\end{equation*}
The third derivative is:
\begin{equation*}
g'''(x)=6x^{4}=\frac{6}{x^4}.
\end{equation*}
The fourth derivative is:
\begin{equation*}
g^{(4)}(x)=24x^{5}=\frac{24}{x^5}.
\end{equation*}
Note that unlike the polynomial in Example2.73, the function in Example2.74 has as many derivatives as we like. They never go to 0!
Example2.75
Consider the function \(h(t)=\sqrt{t^2+4}\text{,}\) find the \(3rd\) derivative of \(h(t)\text{.}\)
Start with the first derivative of \(h(t)\text{;}\) in this case will have to use the chain rule:
\begin{equation*}
\frac{dh}{dt}=\frac{1}{2}(t^2+4)^{1/2}(2t)
\end{equation*}
To find the second derivative, we need to use both product rule and chain rule:
\begin{equation*}
\frac{d^2h}{dt^2}=\frac{1}{4}(t^2+4)^{3/2}(2t)(2t)+\frac{1}{2}(t^2+4)^{1/2}(2)=(t^2+4)^{3/2}(t^2)+(t^2+4)^{1/2}.
\end{equation*}
To take the third derivative, we will have to use the product rule on the first term and the chain twice:
\begin{equation*}
\frac{d^3h}{dt^3}=\frac{3}{2}(t^2+4)^{5/2}(2t)(t^2)+(t^2+4)^{3/2}(2t)+\frac{1}{2}(t^2+4)^{3/2}(2t)
\end{equation*}
Note that when you take higher order derivatives involving the chain rule you must be very careful to keep track of every term.