Differentiation
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Transcript of Differentiation
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Introduction to Differentiation
Shirleen Stibbe www.shirleenstibbe.co.uk
M203 Pure Mathematics Summerschool
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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) =
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)(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)
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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!
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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
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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.
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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
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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
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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)
+
-
+ -
+
↗ ↗ ↘