*Differentiation* is a method used to compute the rate of change of a function $f(x)$ with respect to its input $x$. This rate of change is known as the *derivative* of $f$ with respect to $x$.

The first derivative of a function $y=f(x)$ is denoted $\dfrac{\mathrm{d}y}{\mathrm{d}x}$, where $\mathrm{d}y$ denotes an infinitesimally small change in $y$, and $\mathrm{d}x$ an infinitesimally small change in $x$. It is defined by:

\[\frac{\mathrm{d}y}{\mathrm{d}x}=\lim_{\large{h\to 0}}\left[\frac{f(x+h)-f(x)}{h}\right].\]

The process of finding the derivative by taking this limit is known as differentiation from first principles. In practice it is often not convenient to use this method; the derivatives of many functions can be found using standard derivatives in conjunction with rules such as the chain rule, product rule and quotient rule.

There are several different notations for differentiation. All of them are 'correct', and some are more prevalent in certain fields than others. It's helpful to be aware of each system of notation, even if you don't use it regularly.

Given a function $y=f(x)$, the first derivative of $y$ with respect to $x$ is denoted by \[\dfrac{\mathrm{d}y}{\mathrm{d}x}\]

The second derivative of $y$ with respect to $x$ is denoted by

\[\frac{ \mathrm{d} }{ \mathrm{d}x } \left[ \frac{ \mathrm{d}y }{ \mathrm{d}x } \right] = \frac{ \mathrm{d}^2 y }{ \mathrm{d}x^2 }\]

The $n$^{th} derivative of $y$ with respect to $x$ is denoted by

\[\dfrac{\mathrm{d}^n y}{\mathrm{d}x^n}\]

The derivative of $y$ with respect to $x$ at the point $x=a$ is denoted by

\[\dfrac{\mathrm{d}y}{\mathrm{d}x}\Biggl\vert_{ \Large{x=a} }\]

Also referred to as *prime notation*. In this notation a prime (dash) symbol is used to denote the derivative of a function.

Given a function $y=f(x)$, the first derivative of $y$ with respect to $x$ is denoted by

\[y'(x)=y' \text{ or equivalently } f'(x)=f'.\]

**Note:** It is not necessary to include the argument of the function when using this notation. It should be obvious from context what the argument of the function is. If it's not clear, use a different notation.

The second and third derivatives of $y$ with respect to $x$ are denoted by

\[(y')'=y'' \text{ and } (y'')'=y''' \text{ or equivalently } (f')'=f'' \text{ and } (f'')'=f'''.\]

A superscript Roman numeral or bracketed number is used to denote a higher derivative. For example, the fourth derivative of $y$ with respect to $x$ is denoted

\[y^{ \textrm{iv} } \text{ or } y^{ (4) }.\]

The $n$^{th} derivative of $y$ with respect to $x$ is denoted by

\[y^{(n)} \text{ or equivalently } f^{(n)}.\]

Also referred to as *dot notation*. In this notation a dot is placed above the function name to denote the time derivative of a function.

Given a function $x=f(t)$, the first and second derivatives of $x$ with respect to time $t$ are denoted by

\[\dot{x} \text{ and } \ddot{x}.\]

**Note:** this notation is used strictly to denote a derivative with respect to time.

The derivative of a constant is $0$ - a constant function does not change.

Differentiation is a *linear operation*. Given functions $f(x)$ and $g(x)$ and real numbers $\alpha$ and $\beta$, the following property holds:

\[\dfrac{\mathrm{d} }{\mathrm{d}x}\Bigl[\alpha\,f(x)+\beta\;g(x)\Bigl]=\alpha\dfrac{\mathrm{d}f}{\mathrm{d}x}+\beta\dfrac{\mathrm{d}g}{\mathrm{d}x}\]

A table of standard derivatives, where $n$ is any real number and $k,a$ are any constants.

$f(x)$ |
$f'(x)$ |
---|---|

$k$ |
$0$ |

$x^n$ |
$nx^{n-1}$ |

$kx$ |
$k$ |

$\ln{x}$ |
$\frac{1}{x}$ |

$\mathrm{e}^{x}$ |
$\mathrm{e}^{x}$ |

$\mathrm{e}^{u(x)}$ |
$\mathrm{e}^{u(x)}\dfrac{\mathrm{d}u(x)}{\mathrm{d}x}$ |

$a^x$ |
$a^x\ln{a}$ |

$\sin{x}$ |
$\cos{x}$ |

$\cos{x}$ |
$-\sin{x}$ |

$\tan{x}$ |
$\sec^2{x}$ |

Given a function $y=f(x)$, the derivative $f'(x)$ gives the *gradient* of the graph of this function at a point $x$. This is also the gradient of the tangent to the curve at that point.

The gradient of the graph $y=x^2$ at $x=2$ is $\displaystyle\dfrac{\mathrm{d}}{\mathrm{d}y}\bigl[x^2\bigl]\Biggl|_{x=2}= 2x\Biggl|_{x=2}=2\cdot2=4$.

Stationary points of a function $y=f(x)$ are the points where the derivative is zero. They can be found by setting $\dfrac{\mathrm{d}y}{\mathrm{d}x}=0$ and solving for $x$. The nature of these stationary points (**maxima**, **minima** or **points of inflection**) can be determined by the derivative tests.

Since these local maxima and minima represent the maximum or minimum value of $f(x)$ within a local region, finding stationary points is useful in optimisation problems.

**Note:** The **global** maximum or minimum values of a function $f(x)$ are always found at a point where either the derivative $f'(x)=0$ or the derivative does not exist. However, a point $x$ for which $f'(x)=0$ is not necessarily a **global** maximum or minimum value of the function $f(x)$. Such points may represent a local maximum, local minimum or stationary point of inflection; thus, further analysis is required to determine whether or not the point is a **global** maximum or minimum.

Physics often involves studying rates of change and the way different quantities interact with each other; derivatives and differential equations are therefore essential to the mathematical descriptions of many physical quantities and phenomena.

**Example**: The *velocity* of an object is defined to be the rate of change of its position with respect to time. For an object in one-dimensional motion with position described by $x=x(t)$, its velocity $v$ is given by:

\[v=\dot{x}=\dfrac{\mathrm{d}x}{\mathrm{d}t}.\]

Similarly, the *acceleration* of an object is defined to be the rate of change of its velocity with respect to time. For an object in one-dimensional motion with velocity $v$, its acceleration $a$ is given by:

\[a=\ddot{x}=\dfrac{\mathrm{d}v}{\mathrm{d}t}=\dfrac{ \mathrm{d} }{\mathrm{d}t}\left[\dfrac{\mathrm{d}x}{\mathrm{d}t}\right]=\dfrac{\mathrm{d}^2 x}{\mathrm{d}t^2}.\]

Suppose an object moves along a horizontal axis. Its position is described by $x(t)=t^3+5t-1+\cos{t}$. Find the velocity and acceleration of the object at time $t=1$.

Recall that the velocity of a moving body with displacement $x(t)$ is $v=\dot{x}=\dfrac{\mathrm{d}x}{\mathrm{d}t}.$

Here $x(t)=t^3+5t-1+\cos{t}$, so the velocity of the object is:

\[v=\dot{x}=\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[t^3+5t-1+\cos{t}\Bigl].\]

Since differentiation is a linear operation each term can be treated separately.

From the table of standard derivatives, the derivative of a function $x^n$ is $\dfrac{ \mathrm{d} }{\mathrm{d}x}\Bigl[x^{ \large{n} }\Bigl]=nx^{ \large{n-1} }$. Apply this rule with $n=3$ to differentiate $t^3$:

\[\dfrac{ \mathrm{d} }{ \mathrm{d}t }\Bigl[t^3\Bigl]=3t^{ \large{3-1} } = 3t^2.\]

From the table of standard derivatives, the derivative of a function $kx$ is $\dfrac{ \mathrm{d} }{ \mathrm{d}x }\Bigl[kx\Bigl]=k.$ Hence the derivative of $5t$ is:

\[\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[5t\Bigl]=5.\]

This result can also be shown by the linearity property of the derivative and the general result for a derivative of $x^n$ (with $n=1$):

\[\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[5t\Bigl]=5\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[t^1\Bigl]=5t^{ \large{1-1} }=5t^{ \large{0} }=5.\]

By the properties of differentiation, the derivative of any constant $c$ is $\dfrac{ \mathrm{d} }{\mathrm{d}x}\Bigl[c\Bigl]=0$. Hence the derivative of $1$ is:

\[\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[1\Bigl]=0.\]

Finally, from the table of standard derivatives, the derivative of $\cos{t}$ is

\[\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[\cos{t}\Bigl]=-\sin{t}.\]

Thus, the velocity of the object is given by:

\begin{align} v=\dot{x} &= \dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[t^3+5t-1+\cos{t}\Bigl] \\ &=3t^2+5-0-\sin{t} \\ &=3t^2+5-\sin{t}. \end{align}

At time $t=1$, the velocity is:

\begin{align} v(1) &= \dfrac{ \mathrm{d}x }{\mathrm{d}t}\Biggl\vert_{ \Large{t=1} } \\ &= 3\cdot1^2+5-\sin{1} \\ &= 8-\sin{1} \\ &\approx 7.16 \text{ ms⁻¹}. \end{align}

**Note:** When working with physical quantities such as velocity or acceleration it is essential to include units of measurement in the solution.

To find the acceleration of the object, recall that the acceleration of a moving body with velocity $v$ is given by $a=\ddot{x}=\dfrac{\mathrm{d}v}{\mathrm{d}t}=\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}$.

Here $v(t)=3t^2+5-\sin{t}$, so

\[a=\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[3t^2+5-\sin{t}\Bigl].\]

As above, the derivative is found by using the properties of differentiation and the table of standard integrals.

The derivative of $3t^2$ is:

\[\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[3t^2\Bigl]=3\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[t^2\Bigl]=3\cdot2t^{ \large{2-1} }=6t.\]

$5$ is a constant, so its derivative is:

\[\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[5\Bigl]=0.\]

The derivative of $-\sin{t}$ is:

\[\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[-\sin{t}\Bigl]=-\dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[\sin{t}\Bigl]=-\cos{t}.\]

Thus the acceleration of the object is given by:

\begin{align} a=\ddot{x} &= \dfrac{ \mathrm{d} }{\mathrm{d}t}\Bigl[3t^2+5-\sin{t}\Bigl] \\ &= 6t+0-\cos{t} \\ &= 6t-\cos{t}. \end{align}

At time $t=1$, the acceleration is:

\begin{align} a(1) &= \dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}\Biggl\vert_{ \Large{t=1} } \\ &=6\cdot1-\cos{1} \\ &\approx 5.46\text{ ms⁻²}. \end{align}

Prof. Robin Johnson finds the first and second derivatives of $y(x)+x^3-2\mathrm{e}^{2\large{x}}$ and $x(t)=2\cos{(2t)}$.

These workbooks produced by HELM are good revision aids, containing key points for revision and many worked examples.

Test yourself: Numbas test on differentiation

Test yourself: Numbas test on differentiation, including the chain, product and quotient rules

- Basic differentiation - a refresher workbook at
**math**centre. - Table of derivatives leaflet at
**math**centre. - Extending the table of derivatives workbook at
**math**centre. - Facts and Formulae leaflet from
**math**centre has a table of derivatives and the rules for differentiation. - Taking derivatives at Khan Academy