Title: Default for 11.2 Using a Table of Derivatives
Course: HELM Workbooks 2023-2024

Table of derivatives

The table below lists some of the common functions used in engineering and their corresponding derivatives. Remember that in each case the function in the right-hand column gives the rate of change, or the gradient of the graph, of the function on the left at a particular value of \(x\).

N.B. The angle must always be in radians when differentiating trigonometric functions.

Function | Derivative
constant | 0
\(x\) | \(1\)
\(kx\) | \(k\)
\(x^n\) | \(nx^{n-1}\)
\(kx^n\) | \(knx^{n-1}\)
\(e^x\) | \(e^x\)
\(e^{kx}\) | \(ke^{kx}\)
\(\ln x\) | \(1/x\)
\(\ln kx\) | \(1/x\)
\(\sin x\) | \(\cos x\)
\(\sin kx\) | \(k \cos kx\)
\(\sin (kx + c)\) | \(k \cos (kx + c) \)
\(\cos x\) | \(-\sin x\)
\(\cos kx\) | \(-k \sin kx\)
\(\cos (kx + c)\) | \(-k \sin (kx + c) \)
\(\tan x\) | \(\sec^2 x\)
\(\tan kx\) | \(k \sec^2 kx\)
\(\tan (kx + c)\) | \(k \sec^2 (kx + c) \)

Key Point

Particularly important is the rule for differentiating powers of functions:

If \[y = x^n \hspace{2cm} \text{then} \hspace{2cm} \dfrac{dy}{dx} = nx^{n-1} \]
For example, if \(y = x^3\) then \( \dfrac{dy}{dx} = 3x^2\).

Example

Use the table of derivatives to find \( \dfrac{dy}{dx} \) when \(y\) is given by

1. \(7x\)
2. \(14 \)
3. \(5x^2\)
4. \(4x^7\)

Solution

1. We note that \(7x\) is of the form \(kx\) where \(k = 7\). Using the table we then have \( \dfrac{dy}{dx} = 7\).
2. Noting that \(14\) is a constant we see that \( \dfrac{dy}{dx} = 0\).
3. We see that \(5x^2\) is of the form \(kx^n\), with \(k = 5\) and \(n = 2\). The derivative, \(knx^{n-1}\), is then \(10x^1\), or more simply, \(10x\). So if \(y = 5x^2\), then \( \dfrac{dy}{dx} = 10x\).
4. We see that \(4x^7\) is of the form \(kx^n\), with \(k = 4\) and \(n = 7\). Hence the derivative, \( \dfrac{dy}{dx}\), is given by \(28x^6\).

Exercise

Use the table of derivatives to find \( \dfrac{dy}{dx}\) when
1. \(y = [value] \)
Hint: Write \(\sqrt{x}\) as \(x^{1/2}\) , and use the result for differentiating \(x^n\) with \(n = \dfrac{1}{2}\).
\( \dfrac{dy}{dx} \) =
2. \(y =  \displaystyle [value] \)
Hint: Write \( \dfrac{ [value]}{x^3} \) as \( [value]x^{-3}\) and use the result for differentiating \(kx^n\) with \(k = [value]\) and \(n = -3\).
\( \dfrac{dy}{dx} \) =
Feedback: 1. We have \(y = [value] = x^{1/2}, \) so: \[ \dfrac{dy}{dx} = nx^{n-1} = \dfrac{1}{2}x^{1/2 - 1} = \dfrac{1}{2}x^{-1/2} \] This may be written as \( \dfrac{1}{2\sqrt{x}} \). 2. We have \(y = \displaystyle [value] = [value]x^{-3}, \) so: \[ \dfrac{dy}{dx} = [value](-3)x^{-3-1} = - [value]x^{-4} \]

Exercise

Although the derivative table is written using \(x\) as the independent variable, the table can be used for any variable. Use the table of derivatives to find
1. Find \(\dfrac{dz}{dt}\) given \( z = e^t \)
\( \dfrac{dz}{dt} \) =
2. Find \( \dfrac{dp}{dt}\) given \( p = [value] \)
\( \dfrac{dp}{dt} \) =
3. Find \( \dfrac{dz}{dy}\) given \( z = [value] \)
\( \dfrac{dz}{dy} \) =
Feedback: 1. From the Table, if \(y = e^x\) then \( \dfrac{dy}{dx} = e^x \). Hence if \(z = e^t\) then \( \dfrac{dz}{dt} = e^t \). 2. From the Table, if \(y = e^{ [value]x}\) then \( \dfrac{dy}{dx} = [value]e^{ [value]x} \). Hence if \(p = e^{ [value]t}\) then \( \dfrac{dp}{dt} = [value]e^{ [value]t} \). 3. From the Table, if \(y = e^{- [value]x}\) then \( \dfrac{dy}{dx} = -[value]e^{- [value]x} \). Hence if \(z = e^{ -[value]y}\) then \( \dfrac{dz}{dy} = - [value]e^{ -[value]y} \).

Exercise

 Use the table of derivatives to find
1. \(\dfrac{dy}{dt}\) given \( y = [value] \)
\( \dfrac{dy}{dt} \) =
2. \( \dfrac{dy}{dx}\) given \( y =\displaystyle [value] \)
\( \dfrac{dy}{dx} \) =
3. \( \dfrac{dz}{dt}\) given \( z = [value] \)
\( \dfrac{dz}{dt} \) =
Feedback: 1. We have \( y= [value]\) so \( \dfrac{dy}{dt} = [value]. \) 2. We have \( y=\displaystyle [value]\) so \( \dfrac{dy}{dx}= \displaystyle [value]. \) 3. We have \( z = [value] \) so \( \dfrac{dz}{dt} = [value]. \)

Exercise

Find the derivatives of the following functions with respect to \(x\):
1. \([value]\)
2. \([value] \)
3. \([value]\)
4. \([value]\)
Feedback: 1. For \( y= [value]\) the derivative is \( y' = [value]. \) 2. For \( y= [value]\) the derivative is \( y' = [value]. \) 3. For \( y= [value]\) the derivative is \( y' = [value].\) 4. For \( y= [value]\) the derivative is \( y' = [value]. \)

Exercise

Find \(\dfrac{dz}{dt} \)  when \(z\) is given by:
1. \(\displaystyle [value]\)
2. \( [value] \)
3. \([value] t^{-2}\)
4. \(-\dfrac{[value]}{2} t^{ \frac{ [value] }{2} }\)
5. \( [value] \)
Feedback: 1. We can rewrite \( y= \displaystyle[value]\) as \([value] t^{-3}.\) Then, using the \(kx^n\) rule, we have that the derivative is: \( y' = [value] \cdot (-3) \cdot t^{-4} = \displaystyle [value]. \) 2. For \(y = [value]\), we use the \(x^n\) rule to get \( y' = \dfrac{3}{2} t^{\frac{3}{2}-1} = \displaystyle [value] .\) 3. For \( y= [value] t^{-2}\), we use the \(kx^n\) rule. The derivative is \( y' = [value] \cdot (-2) \cdot t^{-2-1} = \displaystyle [value] .\) 4. For \( y= [value] \), we also use the \(kx^n\) rule. The derivative is \( y' = -\frac{[value]}{2} \cdot \frac{[value]}{2} \cdot t^{\frac{[value]}{2} - 1} = \displaystyle[value] .\) 5. For \( y= [value]\), we use the \(\ln kx\) rule, which gives \( y' =\displaystyle [value] .\)

Exercise

Find the derivative of each of the following with respect to the appropriate variable:
1. \([value]\)
2. \([value]\)
3. \([value]\)
4. \([value]\)
5. \( [value]\)
Feedback: 1. For \( y= [value]\) the derivative is \( y' = [value]. \) 2. For \( y= [value]\) the derivative is \( y' = [value]. \) 3. For \( y= [value]\) the derivative is \( y' = [value]. \) 4. For \( y= [value]\) the derivative is \( y' = [value]. \) 5. For \( y= [value]\) the derivative is \( y' = [value]. \)

Exercise

Find the derivative of each of the following with respect to the appropriate variable:
1. \(\sin (-[value]x)\)
2. \(\cos \dfrac{ [value]}{2}t\)
3. \(\tan \pi x\)
4. \(e^{ \frac{x}{[value]} }\)
5. \( \ln \dfrac{ [value] }{3} t \)

Engineering Example 1

Electrostatic potential

Introduction

The electrostatic potential due to a point charge \(Q\) coulombs at a position \(r\) (m) from the charge is given by \[ V = \dfrac{Q}{4 \pi \epsilon_0 r} \] where \( \epsilon_0 \), the permittivity of free space, \( \approx 8.85 \times 10^{-12}\  \mathrm{Fm^{-1}}\) and \(\pi \approx 3.14\).

The field strength at position \(r\) is given by \(E =- \dfrac{dV}{dr} \).

Problem in words

Find the electric field strength at a distance of \(5 \ \mathrm{m}\) from a source with a charge of \(1 \ \mathrm{coulomb}\).

Mathematical statement of the problem

\( V = \dfrac{Q}{4 \pi \epsilon_0 r} \).

Substitute for \( \epsilon_0\) and \(\pi\) and use \(Q = 1\), so that \[ V = \dfrac{1}{4 \times 3.14 \times 8.85 \times 10^{-12} r} \approx \dfrac{9 \times 10^9}{r} = 9 \times 10^9 r^{-1}. \] We need to differentiate \(V\) in order to find the electric field strength from the relationship \( E = - \dfrac{dV}{dr} \)

Mathematical analysis

\( E = - \dfrac{dV}{dr} = -9 \times 10^9(-r^{-2}) = 9 \times 10^9 r^{-2} \)

When \(r = 5\). \[ E = \dfrac{9 \times 10^9}{25} = 3.6 \times 10^8 \ \mathrm{Vm}^{-1} \]

Interpretation

The electric field strength is \( 3.6 \times 10^8 \ \mathrm{Vm}^{-1}\) at \(r = 5 \ \mathrm{m}\).

Note that the field potential varies with the reciprocal of distance (i.e. inverse linear law with distance) whereas the field strength obeys an inverse square law with distance.

Extending the table of derivatives

We now quote simple rules which enable us to extend the range of functions which we can differentiate. The first two rules are for differentiating sums or differences of functions. The reader should note that all of the rules quoted below can be obtained from first principles using the approach outlined previously.

Key Point

Rule 1: The derivative of \(f(x) + g(x)\) is \( \dfrac{df}{dx} + \dfrac{dg}{dx} \) i.e. \[\dfrac{d}{dx} (f(x)+g(x))=\dfrac{df}{dx} + \dfrac{dg}{dx}. \]

Rule 2: The derivative of \(f(x) - g(x)\) is \( \dfrac{df}{dx} - \dfrac{dg}{dx} \) i.e. \[\dfrac{d}{dx} (f(x)-g(x))=\dfrac{df}{dx} - \dfrac{dg}{dx} .\]

These rules say that to find the derivative of the sum (or difference) of two functions, we simply calculate the sum (or difference) of the derivatives of each function.

Example

Find the derivative of \(y = x^6 + x^4\).

Solution

We simply calculate the sum of the derivatives of each separate function: \[ \dfrac{dy}{dx} = 6x^5 + 4x^3\]

The third rule tells us how to differentiate a multiple of a function. We have already met and applied particular cases of this rule which appear in the Table of derivatives.

Key Point

Rule 3: If \(k\) is a constant, the derivative of \(kf(x)\) is \(k \dfrac{df}{dx} \) i.e. \(\dfrac{d}{dx} (kf(x))=k\dfrac{df}{dx} \).

This rule tells us that if a function is multiplied by a constant, \(k\), then the derivative is also multiplied by the same constant, \(k\).

Example

Find the derivative of \(y = 8e^{2x} \).

Solution

Here we are interested in differentiating a multiple of the function \(e^{2x}\). We differentiate \(e^{2x}\), giving \(2e^{2x}\), and multiply the result by \(8\). Thus \[ \dfrac{dy}{dx} = 8 \times 2e^{2x} = 16 e^{2x} \]

Example

Find the derivative of \(y = 6\sin 2x + 3x^2 - 5e^{3x} \).

Solution

We differentiate each part of the function in turn. \begin{align} y &= 6\sin 2x + 3x^2 - 5e^{3x} \\ \dfrac{dy}{dx} &= 6(2\cos 2x) + 3(2x) - 5(3e^{3x}) \\ &= 12\cos 2x + 6x - 15e^{3x} \end{align}

Exercise

Find \(\dfrac{dy}{dx}\) where \( y = [value] \)
1. First find the derivative of \( [value] \)
2. Next find the derivative of \( [value] \)
3. Finally, combine your results to find the derivative of \( [value]\)

\( \dfrac{dy}{dx} =\)
Feedback: We have \( y= [value].\) 1. The derivarive of [value] is [value]. 2. The derivative of [value] is [value]. 3. So the derivative of \(y\) is \(\dfrac{dy}{dx} = [value].\)

Exercise

Find \( \dfrac{dy}{dx} \) when
1. \(y=[value]\)
\( \dfrac{dy}{dx} =\)
2. \( y=[value] \)
\( \dfrac{dy}{dx} =\)
3. \(y= \displaystyle [value]\)
\( \dfrac{dy}{dx} =\)
4. \(y= \displaystyle [value]\)
\( \dfrac{dy}{dx} =\)
5. \( y= [value]\)
\( \dfrac{dy}{dx} =\)
Feedback: 1. For \( y= [value]\) the derivative is \( y' = [value] \). 2. For \( y= [value]\) the derivative is \( y' = [value] \). 3. For \(\displaystyle y= [value]\) the derivative is \(\displaystyle y' = [value] \). 4. We rewrite \(\displaystyle y= [value]\) as \(\displaystyle\frac{x}{2} + \frac{3}{4}\). The derivative is then \(\displaystyle y' = \frac{1}{2} + 0 = [value] \). 5. We expand \( y= [value]\) to [value]. Using the sum rule, the derivative is \( y' = [value] \).

Exercise

Find the derivative of each of the following functions:
1. \( z(t) = [value]\)
\( \dfrac{dz}{dt} =\)
2. \( h(v) = [value] \)
\( \dfrac{dh}{dv} =\)
3. \( m(n) = \displaystyle [value]\)
\( \dfrac{dm}{dn} =\)
4. \( H(t) = \displaystyle [value]\)
\( \dfrac{dH}{dt} =\)
5. \( S(r) = [value] \)
\( \dfrac{dS}{dr} =\)
Feedback: 1. For \( z= [value]\) the derivative is \[ z' = [value]. \] 2. For \( h= [value]\) the derivative is \[ h' = [value]. \] 3. For \(\displaystyle m= [value]\) the derivative is \[ m' = [value]. \] 4. For \(\displaystyle H= [value]\) the derivative is \[ H' = [value]. \] 5. For \(\displaystyle S = [value]\), first we expand the binomial: \[S = [value].\] Then we see that the derivative is \[S' = [value].\]

Exercise

Find the derivative of each of the following functions:
1. \( A(t) = [value]\)
\( A' =\)
2. \( B(s) = \displaystyle [value] \)
\( B' =\)
3. \( V(r) = \displaystyle [value] \)
\( V' =\)
4. \( M(\theta) = [value] \)
\( M' =\)
5. \( H(t) = [value] \)
\( H' =\)
Feedback: 1. For \( A= [value]\), first we expand the binomial: \[ A = [value]. \] Then we see that the derivative is \[ A' = [value]. \] 2. For \(\displaystyle B= [value]\) the derivative is \[ B' = [value]. \] 3. For \(\displaystyle V= [value]\), first we expand the two binomials: \[ V = [value]. \] Then we see that the derivative is \[ V' = [value]. \] 4. For \(\displaystyle M= [value]\) the derivative is \[ M' = [value]. \] 5. For \( H= [value]\) the derivative is \[ H' = [value]. \]

Evaluating a derivative

The need to find the rate of change of a function at a particular point occurs often. We do this by finding the derivative of the function, and then evaluating the derivative at that point. When taking derivatives of trigonometric functions, any angles must be measured in radians. Consider a function, \(y(x)\). We use the notation \(\dfrac{dy}{dx}(a)\) or \(y'(a)\) to denote the derivative of \(y\) evaluated at \(x = a\). So \(y'(0.5)\) means the value of the derivative of \(y\) when \(x = 0.5\).

Example

Find the value of the derivative of \(y = x^3\) where \(x = 2\). Interpret your result.

Solution

We have \(y = x^3\) and so \( \dfrac{dy}{dx} = 3x^2\).

When \(x = 2\), \(\dfrac{dy}{dx} = 3(2)^2 = 12\), that is, \( \dfrac{dy}{dx}(2) = 12\) (Equivalently, \(y'(2) = 12\)).

The derivative is positive when \(x = 2\) and so \(y\) is increasing at this point. When \(x = 2\), \(y\) is increasing at a rate of \(12\) vertical units per horizontal unit.

Engineering Example 2

Electromotive force

Introduction

Potential difference in an electrical circuit is produced by electromotive force (e.m.f) which is measured in volts and describes the force that maintains current flow around a closed path. Every source of continuous electrical energy, including batteries, generators and thermocouples, consist essentially of an energy converter that produces an e.m.f. An electric current always produces a magnetic field. So the current \(i\) which flows round any closed path produces a magnetic flux \(\phi\) which passes through that path. Conversely, if another closed path, i.e. another coil, is placed within the first path, then the magnetic field due to the first circuit can induce an e.m.f and hence a current in the second coil. The simplest closed path is a single loop. More commonly, helical coils, known as search coils, with known area and number of turns, are used. The induced e.m.f depends upon the number of turns in the coil. The search coil is used with a fluxmeter to measure the change of flux linkage.

Problem in words

A current \(i\) is travelling through a single turn loop of radius \(1 \ \mathrm{m}\). A 4-turn search coil of effective area \(0.03 \ \mathrm{m^2}\) is placed inside the loop. The magnetic flux in weber (Wb) linking the search coil is given by: \[ \phi = \mu_0\dfrac{iA}{2r} \] where \(r\) (m) is the radius of the current carrying loop, \(A\) (m^2) is the area of the search coil and \(\mu_0\) is the permeability of free space, \( 4 \pi \times 10^{-7} \ \mathrm{Hm}^{-1}\).

Find the e.m.f. (in volts) induced in the search coil, given by \( \epsilon = -N \left( \dfrac{d\phi}{dt} \right) \) where \(N\) is the number of turns in the search coil, and the current is given by \[ i = 20\sin(20\pi t) + 50 \sin(30\pi t).\]

Mathematical statement of the problem

Substitute \( i = 20 \sin (20 \pi t) + 50 \sin(30 \pi t) \) into \(\phi = \mu_0 \dfrac{iA}{2r} \) and find \(\epsilon = -N \left( \dfrac{d\phi}{dt} \right) \) where \(r = 1 \ \mathrm{m}\), \(A = 0.03 \ \mathrm{m}^2\), \(\mu_0 = 4 \times 10^{-7} \ \mathrm{Hm}^{-1}\) and \(N = 4\).

Mathematical analysis

\[ \phi = \mu_0 \dfrac{iA}{2r} = \dfrac{\mu_0 A}{2r} (20\sin (20 \pi t) + 50 \sin( 30 \pi t)) \] So \begin{align} \dfrac{d\phi}{dt} &= \dfrac{\mu_0 A}{2r} ( 20\pi \times 20\cos(20\pi t) + 30\pi \times 50\cos(30\pi t) ) \\ &= \dfrac{\mu_0 A}{2r} ( 400\pi \cos(20 \pi t) + 1500 \pi \cos(30\pi t) ) \end{align} so \[ \epsilon = -N \dfrac{d\phi}{dt} = -4 \dfrac{\mu_0 A}{2r}( 400\pi \cos(20 \pi t) + 1500 \pi \cos(30 \pi t)).\] Now, \( \mu_0 = 4\pi \times 10^{-7} \), \(A = 0.03\) and \(r = 1\). So \begin{align} \epsilon &= \dfrac{-4 \times 4\pi \times 10^{-7} \times 0.03}{2 \times 1} ( 400\pi \cos(20 \pi t) + 1500 \pi \cos(30\pi t) ) \\ &= -7.5398 \times 10^{-8}(1256.64 \cos(20 \pi t) + 4712.39 \cos(30 \pi t)) \\ &= -9.475 \times 10^{-5} \cos(20 \pi t) + 3.553 \times 10^{-4} \cos(30\pi t) \end{align}

Interpretation

The induced e.m.f. is \( -9.475 \times 10^{-5} \cos(20 \pi t) + 3.553 \times 10^{-4} \cos(30\pi t) \).

The graphs below show the initial current in the single loop and the e.m.f induced in the search coil.

[Image]

Note that the induced e.m.f does not start at zero, which the initial current does, and has a different pattern of variation with time.

Exercise

1. Calculate the derivative of \(y(x) = [value]\) when \(x = [value]\) radians to \(3\) decimal places.
\( \dfrac{dy}{dx}([value]) =\)
2. Calculate the rate of change of \(i(t) = [value]\) when \(t =\displaystyle [value] \) radians
\( \dfrac{di}{dt}\left(\displaystyle [value]\right) =\)
3. Calculate the rate of change of \(F(t) = [value]\) when \(t = [value]\) radians to \(3\) decimal places.
\( \dfrac{dF}{dt}([value]) =\)
Feedback: 1. We have \( y(x) = [value]\) so \[ \dfrac{dy}{dx} = [value],\] which when evaluated at \(x = [value]\) is \([value]. \) 2. We have \(i(t) = [value]\) so \[ \dfrac{di}{dt} = [value],\] which when evaluated at \( t = \displaystyle [value] \) is [value]. 3. We have \(F(t) = [value]\) so \[ \dfrac{dF}{dt}= [value],\] which when evaluated at \(t = [value]\) is [value].