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)
-
+