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Differentiation

Definition

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.

Notation

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.

Leibniz's Notation

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

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

Lagrange's Notation

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

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

Newton's Notation

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.

Properties

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:

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

Table of Derivatives

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

$f(x)$

$f'(x)$

$k$

$0$

$x^n$

$nx^{n-1}$

$kx$

$k$

$\ln{x}$

$\frac{1}{x}$

$\mathrm{e}^{x}$

$\mathrm{e}^{x}$

$\mathrm{e}^{u(x)}$

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

$a^x$

$a^x\ln{a}$

$\sin{x}$

$\cos{x}$

$\cos{x}$

$-\sin{x}$

$\tan{x}$

$\sec^2{x}$

Applications of Differentiation

The gradient of a graph

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.

Example

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 and Optimisation

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

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:

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

Worked Example

Example 1

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 $t=1$.

Solution

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}

Video Example

Example

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

Workbooks

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

Test Yourself

Test yourself: Numbas test on differentiation

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

See Also

External Resources

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