Differentiation

Introduction to Differentiation Shirleen Stibbe

www.shirleenstibbe.co.uk

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M203 Pure Mathematics Summerschool

Slope of the tangent at a point on f y = f(x) f(c+h) f(c+h) - f(c) f(c)

h c

c+h

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

f (c + h) − f (c ) Q(h) = h [called the difference quotient for f at c] As h → 0, chord → tangent to f at c. f is differentiable at c, and f ʹ′(c ) = lim Q(h) h→0

if the limit exists (and is finite).

Left derivative:

fLʹ′ (c ) = lim− Q (h) h →0

Right derivative: fRʹ′ (c ) = lim+ Q (h ) h →0

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?

1)

⎧2 x f ( x ) = ⎨ 1 x2 + 2 ⎩ 2

2)

⎧2 − x f ( x ) = ⎨ ⎩ x

x ≤2 x >2

3)

⎧2 + x f ( x ) = ⎨ 2 ⎩ x

x ≤2 x >2

NB: Draw a picture!

x ≤2 x >2

Rules 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) 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.

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 f (b ) − f (a ) f ʹ′(c ) = b−a y = f(x) f(b)

f(a) a

c

b

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 f (x ) f ʹ′( x ) lim lim exists and equals x →c g ( x ) x →c g ʹ′( x ) 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:

x − 4x 4 + 3x 5 lim x →1 ( x − 1)2 NB: Check criteria at each stage

Some standard derivatives f(x)

You really should know these

Applying the rules:

f'(x)

xn

nxn-1 (n â&#x2030; 0)

sin(x)

cos(x)

cos(x)

-sin(x)

ex

ex

logex

1/x (x > 0)

Product (fg)' = f'g + fg' g = x-2,

f = sin(x),

f' =

g' =

(sin(x)x-2)' = Quotient (f/g)' = (f'g - fg') / g2 f = sin(x),

g = x 2,

f' =

g' =

g2 =

(sin(x)/x2)' = Inverse (f-1)'(d) = 1/f'(c) [ f(c) = d, f monotonic, f'(c) â&#x2030; 0 ] f(c) = ec = d (loged)' =

f'(c) =

=

What differentiation can do for you

f continuous on interval I differentiable on Int I

f(x)

↗

↘

What f'(x) tells you

↗

f'(x) > 0 ⇒ f increasing f'(x) < 0 ⇒ f decreasing f'(c) = 0 ⇒ f(c) a local extremum

+

f'(x)

+

-

What f"(c) tells you f''(c) > 0 ⇒ f(c) a local minimum

+ + f''(c) < 0 ⇒ f(c) a local maximum

--

f''(x)

-

+