Higher order derivatives

Higher order derivatives reveal even deeper insights into the behaviour of functions, such as jerk in motion or the curvature of graphs. These concepts are important in fields like physics, engineering and economics, where understanding complex dynamics is crucial.

Higher order derivatives are derivatives of derivatives, providing insights into how the rate of change itself changes. You may have already encountered second derivatives when looking at Maxima and minima.

Once you reach a derivative that is \(0\), any further derivatives will also be \(0\).

There are a few ways to denote higher order derivatives for a function \(y=f(x)\), as summarised in the table.

Points of inflection

When graphing a function, the second derivative is the rate of change of the derivative function. It describes how much the function curves inwards (its concavity) and indicates whether the slope of the function is increasing or decreasing.

• If the graph of the function is concave upward, the second derivative is positive. This indicates that the slope is increasing.
• If the graph of the function is concave downward, the second derivative is negative. This indicates that the slope is decreasing.
• At a point of inflection, where the curve changes concavity, the second derivative is \(0\).

Example 1 – finding higher order derivatives

Find the \(4^{\textrm{th}}\) derivative of \(f(x)=\sin(2x)\).
\[\begin{align*} f'(x) & = 2\cos(2x)\\
f''(x) & = -4\sin(2x)\\
f'''(x) & = -8\cos(2x)\\
f''''(x) & = 16\sin(2x)
\end{align*}\]

Example 2

If \(f(x)=3x^{4}-5x^{2}+7\), find \(f''(x)\).
\[\begin{align*} f'(x) & = 12x^{3}-10x\\
f''(x) & = 36x^{2}-10
\end{align*}\]

Example 3

Find \(\dfrac{d^{2}y}{dt^{2}}\) if \(y=\log_{e}(1+t^{4})\).

We need to use the chain rule. Let \(u=1+t^{4}\) and \(y=\log_{e}(u)\). The derivatives are:
\[\frac{du}{dt}=4t^{3}\] \[\frac{dy}{du}=\frac{1}{u}\] \[\begin{align*} \frac{dy}{dt} & = \frac{dy}{du}\times\frac{du}{dt}\\
& = \frac{1}{u}\times4t^{3}\quad\textrm{substituting }u=1+t^{4}\\
& = \frac{1}{1+t^{4}}\times4t^{3}\\
& = \frac{4t^{3}}{1+t^{4}}
\end{align*}\]

Now, we can find \(\dfrac{d^{2}y}{dt^{2}}\) using the quotient rule.
\[\begin{align*} \frac{d^{2}y}{dt^{2}} & = \frac{((1+t^{4})\cdot12t^{2})-(4t^{3}\cdot4t^{3})}{(1+t^{4})^{2}}\\
& = \frac{12t^{2}+12t^{6}-16t^{6}}{(1+t^{4})^{2}}\\
& = \frac{12t^{2}-4t^{6}}{(1+t^{4})^{2}}
\end{align*}\]

Example 4

Find the \(x\)-coordinates of the points of inflection for the curve with equation:
\[y = \frac{x^{4}}{12}+\frac{x^{3}}{3}-\frac{3x^{2}}{2}+3x-7\]

For points of inflection, we need to find the second derivative.
\[\begin{align*} y' & = \frac{4x^{3}}{12}+\frac{3x^{2}}{3}-\frac{6x}{2}+3\\
& = \frac{x^{3}}{3}+x^{2}-3x+3\\
y'' & = x^{2}+2x-3
\end{align*}\]

Now, we solve for \(x\) when \(y’’=0\).
\[\begin{align*} x^{2}+2x-3 & = 0\\
(x+3)(x-1) & = 0\\
x=-3 & \textrm{ or }x=1
\end{align*}\]

There are points of inflection at \(x=-3\) and \(x=1\) and this is confirmed by the graph of the function.

Exercise – finding higher order derivatives

1. Find the third derivative of \(y=x^{5}-2x^{2}+8x\).
2. If \(f(x)=e^{x^{3}}\), find \(f''(x)\).
3. Given \(y=t^{4}+\dfrac{1}{t^{4}}\), find \(\dfrac{d^{3}y}{dt^{3}}\).
4. The displacement of a particle at time \(t\) is described by the function \(s(t)=t^{2}-\cos(3t)\). What is the initial acceleration of the particle?
5. Find any points of inflection for the graph of the function \(y=\dfrac{x^{4}}{4}-2x^{3}+6x^{2}+x-10\).

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