Download Differentiation: A Comprehensive Guide for Engineers and more Summaries Calculus in PDF only on Docsity!
Chapter 1
Di↵erentiation
1.1 Introduction
Why di↵erentiation? Well, it is a useful tool because many real-world problems rely on the
rates of change of quantities. For example, speed is the rate of change of distance of a moving object.
Sometimes an engineer will need to look at a graph of, for example, distance vs time. In that case, questions about rate of change become questions about gradients, i.e. slopes of the tangent to a curve.
Slope of the chord PQ
Change in y
Change in x
f (x + x) f (x)
x
and as x! 0, chord! tangent.
Therefore: Slope of the tangent at x
dy
dx
= lim x! 0
f (x + x) f (x)
x
Example 1.1. Use the above definition to di↵erentiate y = f (x) = x^2.
dy
dx
= lim x! 0
(x + x)^2 x^2
x
= lim x! 0
x
(^2) + 2x x + ( x) (^2) x
2
x
= lim x! 0
(2x + x)
= 2x.
1.2 Basic di↵erentiation
Now let’s consider the functions given in Table 1.1. These are the basic building blocks of the many functions an engineer will need to di↵erentiate (chances are you already saw these in A-Level).
Let us start by calculating some basic derivatives...
Example 1.2. Compute
d
dx
(2e
x 3 cos x).
Applying the addition formula (Rule 1 in Table 1.2) yields
d
dx
(2e
x 3 cos x) = 2
d
dx
(e
x ) 3
d
dx
(cos x)
= 2e
x 3( sin x)
= 2e
x
So we can find derivatives for sums of functions. However, if we are handling a product of functions, we need the Product Rule instead:
f (x)
df dx
xn^ nxn ^1
1 0
ln (x) x ^1
ex^ ex
sin (x) cos (x)
cos (x) sin (x)
sinh (x) cosh (x)
cosh (x) sinh (x)
Table 1.1: Table of Basic Derivatives
Applying the Quotient Rule gives
d
dx
x 1
x^2 + 1
x^2 + 1
(^) d dx (x^ ^ 1)^ ^ (x^ ^ 1)^
d dx
x^2 + 1
(x^2 + 1)
2
x^2 + 1
⇥ 1 (x 1) ⇥ 2 x
(x^2 + 1)
2
x^2 + 2x + 1
(x^2 + 1)
Example 1.6 (Di↵erentiate tanh x using the quotient rule).
d
dx
(tanh x) =
d
dx
sinh x
cosh x
cosh x (^) ddx (sinh x) sinh x (^) ddx (cosh x)
cosh
2 x
=
cosh ⇥ cosh x sinh x ⇥ sinh x
cosh^2 x
cosh^2 x sinh^2 x
cosh^2 x
and now using the hyperbolic identity
cosh^2 x sinh^2 x ⌘ 1 ,
this leads to
d
dx
(tanh x) =
cosh
2 x
and since
sech x ⌘
cosh x
=) sech^2 x ⌘
cosh^2 x
this leads to the result d
dx
(tanh x) = sech
2 x.
This looks very similar to the following result...
d
dx
(tan x) = sec^2 x,
which uses the trigonometric functions instead of hyperbolic ones. You will get to prove
this result for yourself in the Problem Sheet!
1.3 The Chain Rule
So far, we have calculated derivatives of sums, products and quotients of functions. But
what happens when you have a function of a function?
Example 1.7. Compute the following derivative
d
dx
(sin 2x).
The Chain Rule says that
d
dx
(f (u(x))) = f
0 (u(x))
du
dx
So we let
u(x) = 2 x,
du
dx
f (u) = sin u
df
du
= cos u
then applying the chain rule gives
d
dx
(sin 2x) =
d
du
(f (u))
du
dx
= 2 cos u,
and rewriting back in terms of the original variable x gives
d
dx
(sin 2x) = 2 cos 2x.
Let’s try another example...
Example 1.8. Compute the following derivative
d
dx
ln
x^2 1
Put
u(x) = x
2 1 , u
0 (x) = 2x,
f (u) = ln u, f 0 (u) =
u
then applying the chain rule gives
d
dx
ln
x^2 1
2 x
u
2 x
x^2 1
You will want to brace yourself for the next example! This one shows you how to use the
chain rule more than once.
Example 1.9. Compute the following derivative
d
dx
sin
ln
x
2 e
x
First apply chain rule with f (u) = sin u, u = ln
x
2 e
x
= cos
ln
x
2 e
x ^ ⇥
d
dx
ln
x
2 e
x
Then apply chain rule again, this time with f (u) = ln u, u = x^2 ex
= cos
ln
x
2 e
x ^1 x^2 ex
d
dx
x
2 e
x
Finally, apply the product rule with u = x^2 , v = ex
= cos
ln
x
2 e
x ^1 x^2 ex
x
2 e
x
x⇤^ .
Logarithmic di↵erentiation
Sometimes it is useful to take logs on both sides of an equation before di↵erentiating. By doing this you are setting up an implicit equation, making this an example of implicit di↵erentiation.
Example 1.13. Di↵erentiate the function y = 10x^ with respect to x.
y = 10x, ) ln y = x ln 10.
and so in di↵erentiating w.r.t x
y
dy
dx
= ln 10,
dy
dx
x ln 10.
Example 1.14. Find d
dx
(xx).
First let y = xx, then ln y = ln xx^ = x ln x.
d
dx
(ln y) =
d
dx
(x ln x)
y
dy
dx
= ln x +
⇢x
⇢x
dy
dx
= y (1 + ln x)
dy
dx
= xx^ (1 + ln x).
Example 1.15.
y =
x^2 cos x
sin 2x
x^2
2 sin x
Take logs and di↵erentiate with respect to x to give
ln y = ln x^2 + ln cos x ln sin 2x 1
y
dy
dx
2 x
x^2
sin x
cos x
cos 2x
sin 2x
dy
dx
= y
x
tan x 2 cot 2x
dy
dx
x^2 cos x
sin 2x
x
tan x 2 cot 2x
Di↵erentiating Inverse functions
Believe it or not, when you di↵erentiate an inverse function, you are using implicit di↵erentiation (again!)
Example 1.16.
Find
dy
dx
when y = sin ^1 x.
y = sin ^1 x
sin y = x
d
dx
(sin y) = 1
cos y
dy
dx
dy
dx
cos y
p 1 x^2
Example 1.17.
Find
dy
dx
when y = cosh ^1 x.
y = cosh ^1 x
x = cosh y
1 = sinh y
dy
dx
(Implicit di↵erentiation)
dy
dx
sinh y
p cosh^2 y 1
(cosh
2 y sinh
2 y ⌘ 1)
p x^2 1
Therefore dy
dx
p x^2 1
1.4 Higher derivatives
Having found
dy
dx
, we can di↵erentiate this again, which gives is the second derivative
d^2 y
dx^2
If we then di↵erentiate again, we get
d^3 y
dx^3
d^4 y
dx^4
, etc. These are collectively known as higher
derivatives.
In general if y = uv then applying the product rule gives:
y(1)^ = u(1)v + uv(1)
y
(2) = u
(2) v + u
(1) v
(1)
(1) v
(1)
(2)
y
(3) = u
(3) v + 3u
(2) v
(1)
(2) v
(1)
(1) v
(2)
= u(3)^ + 3u(2)v(1)^ + 3u(1)v(2)^ + uv(3).
Notice that the binomial coe cients are appearing.
In fact...
y(n)^ = u(n)v +
n
1
u(n 1)v(1)^ +
n
2
u(n 2)v(2)^ + · · ·
n
n 1
u(1)v(n 1)^ + uv(n)
X^ n
k=
n
k
u(n k)v(k), (1.1)
where (^) ✓ n
k
n!
(n k)!k!
Equation 1.1 is known as the Leibniz rule for di↵erentiating a product n times.
Example 1.21.
If y = xex, what is
dny
dxn^
Using the Leibniz rule with v = x, u = ex^ gives
y(n)^ = x
dn
dxn^
(ex) +
n
1
d
dx
(x)
dn ^1
dxn ^1
(ex)
n
2
d^2
dx^2
(x)
dn ^2
dxn ^2
(ex) + 0
= xex^ + n. 1 .ex
= ex(x + n).
Example 1.22.
Let y = x^2 sin x. Find
d^17 y
dx^17
Tip: When applying the Leibniz rule for the function uv you should choose v such that it
becomes zero when di↵erentiated a relatively few number of times (if this is possible). So
we choose u = sin x, v = x^2.
y(17)^ = x^2
d^17
dx^17
(sin x) +
2 x
d^16
dx^16
(sin x)
d^15
dx^15
(sin x) + 0.
