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Introduction to Differentiation

Shirleen Stibbe www.shirleenstibbe.co.uk

M203 Pure Mathematics Summerschool

Slope of the tangent at a point on f

h

f(c+h) - f(c)

c c+h

f(c)

f(c+h)

y = f(x)

As h → 0, chord → tangent to f at c.

Chord from (c, f(c)) to (c+h, f(c+h)) has gradient:

[called the difference quotient for f at c]

)(lim)(0

hQcfh→

=ʹ′f is differentiable at c, and

if the limit exists (and is finite).

hcfhcf )()( −+

Q(h) =

)(lim)(0

hQcfhL −→

=ʹ′Left derivative:

)(lim)(0

hQcfhR +→

=ʹ′Right derivative:

f is differentiable at c if fL'(c) and fR'(c) exist and are equal, and

f'(c) = fL'(c) = fR'(c)

Note: •  f differentiable ⇒ f continuous

•  f discontinuous ⇒ f not differentiable

To show that f is differentiable at c: •  prove that Q(h) → finite limit as h → 0

To show that f is not differentiable at c: •  show that f is not continuous at c •  find a null sequence {an} s.t. Q(an) → ±∞

•  find null sequences {an}, {bn} s.t. Q(an) ≠ Q(bn)

Examples: Are the following functions differentiable at 2?

⎩⎨⎧

>+

≤=

2222

)( 221 xx

xxxf

⎩⎨⎧

>

≤−=

2 22

)(xxxx

xf

⎩⎨⎧

>

≤+=

222

)( 2 xxxx

xf

1)

2)

3)

NB: Draw a picture!

If f is differentiable at c and g is differentiable at f(c) = d, then Composition: (g ○ f)'(c) = g'(f(c))f'(c) Inverse: (f -1)'(d) = 1 / f'(c) if f monotonic, and

f'(x) ≠ 0 in an interval around c.

If g and h are differentiable at c, then so is f, and f'(c) = g'(c) if:

Local rule: f (x) = g(x) on an interval about c Glue rule: f(x) = g(x) if x < c, f(x) = h(x) if x > c,

f(c) = g(c) = h(c), g'(c) = h'(c)

If f and g are differentiable at c, then:

Sum: (f + g)'(c) = f'(c) + g'(c)

Multiple: (λf)'(c) = λf'(c), λ ∈ R

Product: (fg)'(c) = f'(c)g(c) + f(c)g'(c) Quotient: (f / g)'(c) = (f'(c)g(c) - f(c)g'(c)) / [g(c)]2

(if g(c) ≠ 0)

Rules

Mean Value Theorem If f is •  continuous on [a, b] and •  differentiable on ]a, b[ then there exists a point c ∈ ]a, b[ such that

abafbfcf

−−

=ʹ′ )()()(

a b

y = f(x)

f(b)

f(a)

c

Example: g is continuous on [1, 4] and differentiable on ]1, 4[.

If g(4) = 2 and -1 ≤ g'(x) ≤ 2 for x ∈ ]1, 4[,

prove that -4 ≤ g(1) ≤ 5.

L'Hôpital's Rule

Let f and g be differentiable on an open interval I containing c, and f(c) = g(c) = 0. Then

)()(limxgxf

cx→ )()(limxgxf

cx ʹ′ʹ′

→exists and equals

provided this last limit exists.

NB: You may have to use this rule more than once for a particular limit.

Example: Determine whether the following limit exists, and if it does exist, find its value:

2

54

1 )1(34lim

−+−

→ xxxx

x

NB: Check criteria at

each stage

f(x) f'(x)

xn nxn-1 (n ≠ 0)

sin(x) cos(x)

cos(x) -sin(x)

ex ex

logex 1/x (x > 0) Applying the rules:

Product (fg)' = f'g + fg'

f = sin(x), g = x-2, f' = g' =

(sin(x)x-2)' =

Quotient (f/g)' = (f'g - fg') / g2

f = sin(x), g = x2, f' = g' = g2 =

(sin(x)/x2)' =

Inverse (f-1)'(d) = 1/f'(c) [ f(c) = d, f monotonic, f'(c) ≠ 0 ]

f(c) = ec = d f'(c) = =

(loged)' =

Some standard derivatives

You really should know

these

What differentiation can do for you

What f'(x) tells you

f'(x) > 0 ⇒ f increasing

f'(x) < 0 ⇒ f decreasing

f'(c) = 0 ⇒ f(c) a local extremum

f continuous on interval I differentiable on Int I

What f"(c) tells you

f''(c) > 0 ⇒ f(c) a local minimum

f''(c) < 0 ⇒ f(c) a local maximum

+ +

- -

f(x)

f'(x)

f''(x)

+

-

+ -

+

↗ ↗ ↘