Lesson 12: Higher Order Derivatives
Higher Order Derivatives
Let \(y = f(x)\). We can define derivatives of various orders beyond just the first derivative.
Definition: The first derivative of \(y\) with respect to \(x\) is
\[y' = f'(x) = \frac{dy}{dx} = \frac{d}{dx}(f(x)) = f^{(1)}(x).\]Definition: The second derivative of \(y\) with respect to \(x\) is
\[y'' = f''(x) = \frac{d^2y}{dx^2} = \frac{d^2}{dx^2}(f(x)) = f^{(2)}(x).\]Definition: The \(n\)th derivative of \(y\) with respect to \(x\) is
\[\frac{d^ny}{dx^n} = \frac{d^n}{dx^n}(f(x)) = f^{(n)}(x).\]Notation: There are several equivalent notations for derivatives. The first derivative can be written as \(y'\), \(f'(x)\), \(\frac{dy}{dx}\), or \(f^{(1)}(x)\). The second derivative can be written as \(y''\), \(f''(x)\), \(\frac{d^2y}{dx^2}\), or \(f^{(2)}(x)\). For higher order derivatives, we use \(f^{(n)}(x)\) or \(\frac{d^n y}{dx^n}\).
Examples
Example 1
Let \(f(x) = 3x^3 + 4x^2 + 5x + 6\). Find \(f'''(x)\). What is \(f'''(0)-f''(0)\)?
Solution:
We need to differentiate three times.
The first derivative is
\[f'(x) = 9x^2 + 8x + 5.\]The second derivative is
\[f''(x) = 18x + 8.\]The third derivative is
\[f'''(x) = 18.\]We see that
\[f'''(0)-f''(0) = 18-(18\cdot 0 +8)=18-8=10.\]Example 2
Let \(f^{(3)}(x) = 8\csc(7x - 2)\). Find \(f^{(4)}(x)\).
Solution:
We are given the third derivative and need to find the fourth derivative. This means we need to differentiate \(f^{(3)}(x)\) once more.
Using the chain rule, recall that \(\frac{d}{dx}[\csc(u)] = -\csc(u)\cot(u) \cdot \frac{du}{dx}\).
Therefore,
\[f^{(4)}(x) = 8 \cdot (-\csc(7x-2)\cot(7x-2)) \cdot 7 = -56\csc(7x-2)\cot(7x-2).\]Acceleration
Let \(s(t)\) denote the position function, where \(t\) represents time. Recall that the velocity function is the first derivative of the position function, i.e. \(v(t) = s'(t)\).
Definition: The acceleration function, denoted by \(a(t)\), is the derivative of the velocity function. Since the velocity function is the first derivative of the position function, the acceleration function is the second derivative of the position function. So,
\[a(t) = v'(t) = s''(t).\]Example 3
A particle is traveling on a straight line with a position function of
\[s(t) = \frac{2}{3}t^3 + 6t^2,\]where \(t\) is time in seconds and \(s(t)\) is position in feet. What is the acceleration when the velocity of the particle is 54 feet per second?
Solution:
First, we find the velocity function by differentiating the position function. We have that
\[v(t) = s'(t) = 2t^2 + 12t.\]Next, we find the acceleration function by differentiating the velocity function. So,
\[a(t) = v'(t) = 4t + 12.\]We need to find the acceleration when the velocity is 54 feet per second. First, we want to find time \(t\) such that \(v(t) = 54\). We see that:
\[54 = 2t^2 + 12t\] \[\begin{align}&\implies 2t^2 + 12t - 54 = 0\\ &\implies t^2 + 6t - 27 = 0\\ &\implies (t - 3)(t + 9) = 0\end{align}\]Therefore, \(t = 3\) or \(t = -9\). Since time cannot be negative in this context, we have \(t = 3\) seconds.
Now we evaluate the acceleration at \(t = 3\):
\[a(3) = 4\cdot 3 + 12 = 12 + 12 = 24 \text{ ft/s}^2\]Practice Problems
Problem 1
Find the second derivative of \(f(x) = 3x^2\ln(10x)\).
Solution:
We need to find \(f''(x)\). First, we find the first derivative using the product and chain rule:
\[\begin{align}f'(x) &= 6x\ln(10x) + 3x^2 \cdot \frac{1}{10x} \cdot 10\\ &= 6x\ln(10x) + \frac{3x^2}{x}\\ &= 6x\ln(10x) + 3x\end{align}\]Now, we find the second derivative using the product and chain rule (again):
\[\begin{align}f''(x) &= 6\ln(10x) + 6x \cdot \frac{1}{10x} \cdot 10 + 3\\ &= 6\ln(10x) + \frac{6x}{x} + 3\\ &= 6\ln(10x) + 6 + 3\\ &= 6\ln(10x) + 9\end{align}\]Problem 2
Find the second derivative of \(\displaystyle f(x) = \frac{x}{x-1}\).
Solution:
First, we find the first derivative using the quotient rule:
\[f'(x) = \frac{1\cdot(x-1) - 1\cdot x}{(x-1)^2} = \frac{x - 1 - x}{(x-1)^2} = \frac{-1}{(x-1)^2}\]We can rewrite this as:
\[f'(x) = -(x-1)^{-2}\]Now, we find the second derivative using the chain rule:
\[f''(x) = (-1)\cdot (-2)\cdot (x-1)^{-3}\cdot 1 = \frac{2}{(x-1)^3}\]Alternatively, using the quotient rule on \(f'(x) = \frac{-1}{(x-1)^2}\):
\[f''(x) =\frac{0\cdot(x-1)^2 - 2\cdot (x-1) \cdot 1\cdot (-1)}{(x-1)^4} = \frac{2(x-1)}{(x-1)^4} = \frac{2}{(x-1)^3}\]Problem 3
Suppose \(\displaystyle f(x) = 4\sqrt{x} + \frac{8}{x}\). Find \(f''(1)\).
Solution:
We can rewrite the function as
\[f(x) = 4x^{1/2} + 8x^{-1}.\]The first derivative is
\[f'(x) = 4 \cdot \frac{1}{2}\cdot x^{-1/2} + 8\cdot(-1)\cdot x^{-2}= 2x^{-1/2} - 8x^{-2}.\]The second derivative is
\[f''(x) = 2 \cdot \left(-\frac{1}{2}\right)\cdot x^{-3/2} - 8\cdot(-2)\cdot x^{-3}= -x^{-3/2} + 16x^{-3}.\]So, we have that
\[f''(1) = -(1)^{-3/2} + 16\cdot (1)^{-3} = -1 + 16 = 15.\]