lesson 8: basic differentation rules (slides)
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... NYUMathematics.
Sec on 2.3Basic Differenta on Rules
V63.0121.011: Calculus IProfessor Ma hew Leingang
New York University
Announcements
I Quiz 1 this week on1.1–1.4
I Quiz 2 March 3/4 on 1.5,1.6, 2.1, 2.2, 2.3
I Midterm Monday March7 in class
ObjectivesI Understand and usethese differen a onrules:
I the deriva ve of aconstant func on (zero);
I the Constant Mul pleRule;
I the Sum Rule;I the Difference Rule;I the deriva ves of sine
and cosine.
Recall: the derivative
Defini onLet f be a func on and a a point in the domain of f. If the limit
f′(a) = limh→0
f(a+ h)− f(a)h
= limx→a
f(x)− f(a)x− a
exists, the func on is said to be differen able at a and f′(a) is thederiva ve of f at a.
The deriva ve …I …measures the slope of the line through (a, f(a)) tangent tothe curve y = f(x);
I …represents the instantaneous rate of change of f at aI …produces the best possible linear approxima on to f near a.
NotationNewtonian nota on Leibnizian nota on
f′(x) y′(x) y′dydx
ddx
f(x)dfdx
Link between the notations
f′(x) = lim∆x→0
f(x+∆x)− f(x)∆x
= lim∆x→0
∆y∆x
=dydx
I Leibniz thought ofdydx
as a quo ent of “infinitesimals”
I We think ofdydx
as represen ng a limit of (finite) differencequo ents, not as an actual frac on itself.
I The nota on suggests things which are true even though theydon’t follow from the nota on per se
OutlineDeriva ves so far
Deriva ves of power func ons by handThe Power Rule
Deriva ves of polynomialsThe Power Rule for whole number powersThe Power Rule for constantsThe Sum RuleThe Constant Mul ple Rule
Deriva ves of sine and cosine
Derivative of the squaring functionExample
Suppose f(x) = x2. Use the defini on of deriva ve to find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)2 − x2
h
= limh→0
��x2 + 2xh+ h2 −��x2
h= lim
h→0
2x��h+ h�2
��h= lim
h→0(2x+ h) = 2x.
Derivative of the squaring functionExample
Suppose f(x) = x2. Use the defini on of deriva ve to find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)2 − x2
h
= limh→0
��x2 + 2xh+ h2 −��x2
h= lim
h→0
2x��h+ h�2
��h= lim
h→0(2x+ h) = 2x.
Derivative of the squaring functionExample
Suppose f(x) = x2. Use the defini on of deriva ve to find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)2 − x2
h
= limh→0
��x2 + 2xh+ h2 −��x2
h= lim
h→0
2x��h+ h�2
��h= lim
h→0(2x+ h) = 2x.
Derivative of the squaring functionExample
Suppose f(x) = x2. Use the defini on of deriva ve to find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)2 − x2
h
= limh→0
��x2 + 2xh+ h2 −��x2
h
= limh→0
2x��h+ h�2
��h= lim
h→0(2x+ h) = 2x.
Derivative of the squaring functionExample
Suppose f(x) = x2. Use the defini on of deriva ve to find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)2 − x2
h
= limh→0
��x2 + 2xh+ h2 −��x2
h= lim
h→0
2x��h+ h�2
��h
= limh→0
(2x+ h) = 2x.
Derivative of the squaring functionExample
Suppose f(x) = x2. Use the defini on of deriva ve to find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)2 − x2
h
= limh→0
��x2 + 2xh+ h2 −��x2
h= lim
h→0
2x��h+ h�2
��h= lim
h→0(2x+ h) = 2x.
The second derivative
If f is a func on, so is f′, and we can seek its deriva ve.
f′′ = (f′)′
It measures the rate of change of the rate of change!
Leibniziannota on:
d2ydx2
d2
dx2f(x)
d2fdx2
The second derivative
If f is a func on, so is f′, and we can seek its deriva ve.
f′′ = (f′)′
It measures the rate of change of the rate of change! Leibniziannota on:
d2ydx2
d2
dx2f(x)
d2fdx2
The squaring function and its derivatives
.. x.
y
. f.
f′
.
f′′ I f increasing =⇒ f′ ≥ 0I f decreasing =⇒ f′ ≤ 0I horizontal tangent at 0=⇒ f′(0) = 0
The squaring function and its derivatives
.. x.
y
. f
.
f′
.
f′′ I f increasing =⇒ f′ ≥ 0I f decreasing =⇒ f′ ≤ 0I horizontal tangent at 0=⇒ f′(0) = 0
The squaring function and its derivatives
.. x.
y
. f.
f′
.
f′′
I f increasing =⇒ f′ ≥ 0I f decreasing =⇒ f′ ≤ 0I horizontal tangent at 0=⇒ f′(0) = 0
The squaring function and its derivatives
.. x.
y
. f.
f′
.
f′′ I f increasing =⇒ f′ ≥ 0I f decreasing =⇒ f′ ≤ 0I horizontal tangent at 0=⇒ f′(0) = 0
Derivative of the cubing functionExample
Suppose f(x) = x3. Use the defini on of deriva ve to find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)3 − x3
h
= limh→0
��x3 + 3x2h+ 3xh2 + h3 −��x3
h= lim
h→0
3x2��h+ 3xh���1
2 + h���2
3
��h= lim
h→0
(3x2 + 3xh+ h2
)= 3x2.
Derivative of the cubing functionExample
Suppose f(x) = x3. Use the defini on of deriva ve to find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)3 − x3
h
= limh→0
��x3 + 3x2h+ 3xh2 + h3 −��x3
h= lim
h→0
3x2��h+ 3xh���1
2 + h���2
3
��h= lim
h→0
(3x2 + 3xh+ h2
)= 3x2.
Derivative of the cubing functionExample
Suppose f(x) = x3. Use the defini on of deriva ve to find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)3 − x3
h
= limh→0
��x3 + 3x2h+ 3xh2 + h3 −��x3
h
= limh→0
3x2��h+ 3xh���1
2 + h���2
3
��h= lim
h→0
(3x2 + 3xh+ h2
)= 3x2.
Derivative of the cubing functionExample
Suppose f(x) = x3. Use the defini on of deriva ve to find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)3 − x3
h
= limh→0
��x3 + 3x2h+ 3xh2 + h3 −��x3
h= lim
h→0
3x2��h+ 3xh���1
2 + h���2
3
��h
= limh→0
(3x2 + 3xh+ h2
)= 3x2.
Derivative of the cubing functionExample
Suppose f(x) = x3. Use the defini on of deriva ve to find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)3 − x3
h
= limh→0
��x3 + 3x2h+ 3xh2 + h3 −��x3
h= lim
h→0
3x2��h+ 3xh���1
2 + h���2
3
��h= lim
h→0
(3x2 + 3xh+ h2
)= 3x2.
The cubing function and its derivatives
.. x.
y
.f
.
f′
.
f′′
I No ce that f is increasing,and f′ > 0 exceptf′(0) = 0
I No ce also that thetangent line to the graphof f at (0, 0) crosses thegraph (contrary to apopular “defini on” ofthe tangent line)
The cubing function and its derivatives
.. x.
y
.f
.
f′
.
f′′
I No ce that f is increasing,and f′ > 0 exceptf′(0) = 0
I No ce also that thetangent line to the graphof f at (0, 0) crosses thegraph (contrary to apopular “defini on” ofthe tangent line)
The cubing function and its derivatives
.. x.
y
.f
.
f′
.
f′′
I No ce that f is increasing,and f′ > 0 exceptf′(0) = 0
I No ce also that thetangent line to the graphof f at (0, 0) crosses thegraph (contrary to apopular “defini on” ofthe tangent line)
The cubing function and its derivatives
.. x.
y
.f
.
f′
.
f′′
I No ce that f is increasing,and f′ > 0 exceptf′(0) = 0
I No ce also that thetangent line to the graphof f at (0, 0) crosses thegraph (contrary to apopular “defini on” ofthe tangent line)
The cubing function and its derivatives
.. x.
y
.f
.
f′
.
f′′
I No ce that f is increasing,and f′ > 0 exceptf′(0) = 0
I No ce also that thetangent line to the graphof f at (0, 0) crosses thegraph (contrary to apopular “defini on” ofthe tangent line)
Derivative of the square rootExample
Suppose f(x) =√x = x1/2. Fnd f′(x) with the defini on.
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
√x+ h−
√x
h
= limh→0
√x+ h−
√x
h·√x+ h+
√x√
x+ h+√x
= limh→0
(�x+ h)− �xh(√
x+ h+√x) = lim
h→0
��h��h(√
x+ h+√x) =
12√x
Derivative of the square rootExample
Suppose f(x) =√x = x1/2. Fnd f′(x) with the defini on.
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
√x+ h−
√x
h
= limh→0
√x+ h−
√x
h·√x+ h+
√x√
x+ h+√x
= limh→0
(�x+ h)− �xh(√
x+ h+√x) = lim
h→0
��h��h(√
x+ h+√x) =
12√x
Derivative of the square rootExample
Suppose f(x) =√x = x1/2. Fnd f′(x) with the defini on.
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
√x+ h−
√x
h
= limh→0
√x+ h−
√x
h·√x+ h+
√x√
x+ h+√x
= limh→0
(�x+ h)− �xh(√
x+ h+√x) = lim
h→0
��h��h(√
x+ h+√x) =
12√x
Derivative of the square rootExample
Suppose f(x) =√x = x1/2. Fnd f′(x) with the defini on.
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
√x+ h−
√x
h
= limh→0
√x+ h−
√x
h·√x+ h+
√x√
x+ h+√x
= limh→0
(�x+ h)− �xh(√
x+ h+√x)
= limh→0
��h��h(√
x+ h+√x) =
12√x
Derivative of the square rootExample
Suppose f(x) =√x = x1/2. Fnd f′(x) with the defini on.
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
√x+ h−
√x
h
= limh→0
√x+ h−
√x
h·√x+ h+
√x√
x+ h+√x
= limh→0
(�x+ h)− �xh(√
x+ h+√x) = lim
h→0
��h��h(√
x+ h+√x)
=1
2√x
Derivative of the square rootExample
Suppose f(x) =√x = x1/2. Fnd f′(x) with the defini on.
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
√x+ h−
√x
h
= limh→0
√x+ h−
√x
h·√x+ h+
√x√
x+ h+√x
= limh→0
(�x+ h)− �xh(√
x+ h+√x) = lim
h→0
��h��h(√
x+ h+√x) =
12√x
The square root and its derivatives
.. x.
y
I Here limx→0+
f′(x) = ∞ and fis not differen able at 0
I No ce also limx→∞
f′(x) = 0
The square root and its derivatives
.. x.
y
.
f
I Here limx→0+
f′(x) = ∞ and fis not differen able at 0
I No ce also limx→∞
f′(x) = 0
The square root and its derivatives
.. x.
y
.
f
. f′I Here lim
x→0+f′(x) = ∞ and f
is not differen able at 0
I No ce also limx→∞
f′(x) = 0
The square root and its derivatives
.. x.
y
.
f
. f′I Here lim
x→0+f′(x) = ∞ and f
is not differen able at 0I No ce also lim
x→∞f′(x) = 0
Derivative of the cube rootExample
Suppose f(x) = 3√x = x1/3. Find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)1/3 − x1/3
h
= limh→0
(x+ h)1/3 − x1/3
h· (x+ h)2/3 + (x+ h)1/3x1/3 + x2/3
(x+ h)2/3 + (x+ h)1/3x1/3 + x2/3
= limh→0
(�x+ h)− �xh ((x+ h)2/3 + (x+ h)1/3x1/3 + x2/3)
= limh→0
��h��h ((x+ h)2/3 + (x+ h)1/3x1/3 + x2/3)
=1
3x2/3
Derivative of the cube rootExample
Suppose f(x) = 3√x = x1/3. Find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)1/3 − x1/3
h
= limh→0
(x+ h)1/3 − x1/3
h· (x+ h)2/3 + (x+ h)1/3x1/3 + x2/3
(x+ h)2/3 + (x+ h)1/3x1/3 + x2/3
= limh→0
(�x+ h)− �xh ((x+ h)2/3 + (x+ h)1/3x1/3 + x2/3)
= limh→0
��h��h ((x+ h)2/3 + (x+ h)1/3x1/3 + x2/3)
=1
3x2/3
Derivative of the cube rootExample
Suppose f(x) = 3√x = x1/3. Find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)1/3 − x1/3
h
= limh→0
(x+ h)1/3 − x1/3
h· (x+ h)2/3 + (x+ h)1/3x1/3 + x2/3
(x+ h)2/3 + (x+ h)1/3x1/3 + x2/3
= limh→0
(�x+ h)− �xh ((x+ h)2/3 + (x+ h)1/3x1/3 + x2/3)
= limh→0
��h��h ((x+ h)2/3 + (x+ h)1/3x1/3 + x2/3)
=1
3x2/3
Derivative of the cube rootExample
Suppose f(x) = 3√x = x1/3. Find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)1/3 − x1/3
h
= limh→0
(x+ h)1/3 − x1/3
h· (x+ h)2/3 + (x+ h)1/3x1/3 + x2/3
(x+ h)2/3 + (x+ h)1/3x1/3 + x2/3
= limh→0
(�x+ h)− �xh ((x+ h)2/3 + (x+ h)1/3x1/3 + x2/3)
= limh→0
��h��h ((x+ h)2/3 + (x+ h)1/3x1/3 + x2/3)
=1
3x2/3
Derivative of the cube rootExample
Suppose f(x) = 3√x = x1/3. Find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)1/3 − x1/3
h
= limh→0
(x+ h)1/3 − x1/3
h· (x+ h)2/3 + (x+ h)1/3x1/3 + x2/3
(x+ h)2/3 + (x+ h)1/3x1/3 + x2/3
= limh→0
(�x+ h)− �xh ((x+ h)2/3 + (x+ h)1/3x1/3 + x2/3)
= limh→0
��h��h ((x+ h)2/3 + (x+ h)1/3x1/3 + x2/3)
=1
3x2/3
Derivative of the cube rootExample
Suppose f(x) = 3√x = x1/3. Find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)1/3 − x1/3
h
= limh→0
(x+ h)1/3 − x1/3
h· (x+ h)2/3 + (x+ h)1/3x1/3 + x2/3
(x+ h)2/3 + (x+ h)1/3x1/3 + x2/3
= limh→0
(�x+ h)− �xh ((x+ h)2/3 + (x+ h)1/3x1/3 + x2/3)
= limh→0
��h��h ((x+ h)2/3 + (x+ h)1/3x1/3 + x2/3)
=1
3x2/3
The cube root and its derivative
.. x.
y
I Here limx→0
f′(x) = ∞ and fis not differen able at 0
I No ce alsolim
x→±∞f′(x) = 0
The cube root and its derivative
.. x.
y
.
f
I Here limx→0
f′(x) = ∞ and fis not differen able at 0
I No ce alsolim
x→±∞f′(x) = 0
The cube root and its derivative
.. x.
y
.
f
. f′
I Here limx→0
f′(x) = ∞ and fis not differen able at 0
I No ce alsolim
x→±∞f′(x) = 0
The cube root and its derivative
.. x.
y
.
f
. f′
I Here limx→0
f′(x) = ∞ and fis not differen able at 0
I No ce alsolim
x→±∞f′(x) = 0
One moreExample
Suppose f(x) = x2/3. Use the defini on of deriva ve to find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)2/3 − x2/3
h
= limh→0
(x+ h)1/3 − x1/3
h·((x+ h)1/3 + x1/3
)= 1
3x−2/3
(2x1/3
)= 2
3x−1/3
One moreExample
Suppose f(x) = x2/3. Use the defini on of deriva ve to find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)2/3 − x2/3
h
= limh→0
(x+ h)1/3 − x1/3
h·((x+ h)1/3 + x1/3
)= 1
3x−2/3
(2x1/3
)= 2
3x−1/3
One moreExample
Suppose f(x) = x2/3. Use the defini on of deriva ve to find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)2/3 − x2/3
h
= limh→0
(x+ h)1/3 − x1/3
h·((x+ h)1/3 + x1/3
)
= 13x
−2/3(2x1/3
)= 2
3x−1/3
One moreExample
Suppose f(x) = x2/3. Use the defini on of deriva ve to find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)2/3 − x2/3
h
= limh→0
(x+ h)1/3 − x1/3
h·((x+ h)1/3 + x1/3
)= 1
3x−2/3
(2x1/3
)
= 23x
−1/3
One moreExample
Suppose f(x) = x2/3. Use the defini on of deriva ve to find f′(x).
Solu on
f′(x) = limh→0
f(x+ h)− f(x)h
= limh→0
(x+ h)2/3 − x2/3
h
= limh→0
(x+ h)1/3 − x1/3
h·((x+ h)1/3 + x1/3
)= 1
3x−2/3
(2x1/3
)= 2
3x−1/3
x 7→ x2/3 and its derivative
.. x.
y
. f′
I f is not differen able at 0and lim
x→0±f′(x) = ±∞
I No ce alsolim
x→±∞f′(x) = 0
x 7→ x2/3 and its derivative
.. x.
y
.
f
. f′
I f is not differen able at 0and lim
x→0±f′(x) = ±∞
I No ce alsolim
x→±∞f′(x) = 0
x 7→ x2/3 and its derivative
.. x.
y
.
f
. f′
I f is not differen able at 0and lim
x→0±f′(x) = ±∞
I No ce alsolim
x→±∞f′(x) = 0
x 7→ x2/3 and its derivative
.. x.
y
.
f
. f′
I f is not differen able at 0and lim
x→0±f′(x) = ±∞
I No ce alsolim
x→±∞f′(x) = 0
Recap
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I The power goes down byone in each deriva ve
I The coefficient in thederiva ve is the power ofthe original func on
Recap
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I The power goes down byone in each deriva ve
I The coefficient in thederiva ve is the power ofthe original func on
Recap
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I The power goes down byone in each deriva ve
I The coefficient in thederiva ve is the power ofthe original func on
Recap
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I The power goes down byone in each deriva ve
I The coefficient in thederiva ve is the power ofthe original func on
Recap
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I The power goes down byone in each deriva ve
I The coefficient in thederiva ve is the power ofthe original func on
Recap
y y′
x2 2x1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I The power goes down byone in each deriva ve
I The coefficient in thederiva ve is the power ofthe original func on
Recap: The Tower of Power
y y′
x2 2x1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I The power goes down byone in each deriva ve
I The coefficient in thederiva ve is the power ofthe original func on
Recap: The Tower of Power
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I The power goes down byone in each deriva ve
I The coefficient in thederiva ve is the power ofthe original func on
Recap: The Tower of Power
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I The power goes down byone in each deriva ve
I The coefficient in thederiva ve is the power ofthe original func on
Recap: The Tower of Power
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I The power goes down byone in each deriva ve
I The coefficient in thederiva ve is the power ofthe original func on
Recap: The Tower of Power
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I The power goes down byone in each deriva ve
I The coefficient in thederiva ve is the power ofthe original func on
Recap: The Tower of Power
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I The power goes down byone in each deriva ve
I The coefficient in thederiva ve is the power ofthe original func on
Recap: The Tower of Power
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I The power goes down byone in each deriva ve
I The coefficient in thederiva ve is the power ofthe original func on
The Power RuleThere is moun ng evidence for
Theorem (The Power Rule)
Let r be a real number and f(x) = xr. Then
f′(x) = rxr−1
as long as the expression on the right-hand side is defined.
I Perhaps the most famous rule in calculusI We will assume it as of todayI We will prove it many ways for many different r.
The other Tower of Power
OutlineDeriva ves so far
Deriva ves of power func ons by handThe Power Rule
Deriva ves of polynomialsThe Power Rule for whole number powersThe Power Rule for constantsThe Sum RuleThe Constant Mul ple Rule
Deriva ves of sine and cosine
Remember your algebraFactLet n be a posi ve whole number. Then
(x+ h)n = xn + nxn−1h+ (stuff with at least two hs in it)
Proof.We have
(x+ h)n = (x+ h) · (x+ h) · (x+ h) · · · (x+ h)︸ ︷︷ ︸n copies
=n∑
k=0
ckxkhn−k
Remember your algebraFactLet n be a posi ve whole number. Then
(x+ h)n = xn + nxn−1h+ (stuff with at least two hs in it)
Proof.We have
(x+ h)n = (x+ h) · (x+ h) · (x+ h) · · · (x+ h)︸ ︷︷ ︸n copies
=n∑
k=0
ckxkhn−k
Remember your algebraFactLet n be a posi ve whole number. Then
(x+ h)n = xn + nxn−1h+ (stuff with at least two hs in it)
Proof.We have
(x+ h)n = (x+ h) · (x+ h) · (x+ h) · · · (x+ h)︸ ︷︷ ︸n copies
=n∑
k=0
ckxkhn−k
Remember your algebraFactLet n be a posi ve whole number. Then
(x+ h)n = xn + nxn−1h+ (stuff with at least two hs in it)
Proof.We have
(x+ h)n = (x+ h) · (x+ h) · (x+ h) · · · (x+ h)︸ ︷︷ ︸n copies
=n∑
k=0
ckxkhn−k
Remember your algebraFactLet n be a posi ve whole number. Then
(x+ h)n = xn + nxn−1h+ (stuff with at least two hs in it)
Proof.We have
(x+ h)n = (x+ h) · (x+ h) · (x+ h) · · · (x+ h)︸ ︷︷ ︸n copies
=n∑
k=0
ckxkhn−k
Remember your algebraFactLet n be a posi ve whole number. Then
(x+ h)n = xn + nxn−1h+ (stuff with at least two hs in it)
Proof.We have
(x+ h)n = (x+ h) · (x+ h) · (x+ h) · · · (x+ h)︸ ︷︷ ︸n copies
=n∑
k=0
ckxkhn−k
Remember your algebraFactLet n be a posi ve whole number. Then
(x+ h)n = xn + nxn−1h+ (stuff with at least two hs in it)
Proof.We have
(x+ h)n = (x+ h) · (x+ h) · (x+ h) · · · (x+ h)︸ ︷︷ ︸n copies
=n∑
k=0
ckxkhn−k
Pascal’s Triangle..1.
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(x+ h)0 = 1(x+ h)1 = 1x+ 1h(x+ h)2 = 1x2 + 2xh+ 1h2
(x+ h)3 = 1x3 + 3x2h+ 3xh2 + 1h3
. . . . . .
Pascal’s Triangle..1.
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(x+ h)0 = 1(x+ h)1 = 1x+ 1h(x+ h)2 = 1x2 + 2xh+ 1h2
(x+ h)3 = 1x3 + 3x2h+ 3xh2 + 1h3
. . . . . .
Pascal’s Triangle..1.
1
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1
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2
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(x+ h)0 = 1(x+ h)1 = 1x+ 1h(x+ h)2 = 1x2 + 2xh+ 1h2
(x+ h)3 = 1x3 + 3x2h+ 3xh2 + 1h3
. . . . . .
Pascal’s Triangle..1.
1
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1
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2
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(x+ h)0 = 1(x+ h)1 = 1x+ 1h(x+ h)2 = 1x2 + 2xh+ 1h2
(x+ h)3 = 1x3 + 3x2h+ 3xh2 + 1h3
. . . . . .
Proving the Power RuleTheorem (The Power Rule)
Let n be a posi ve whole number. Thenddx
xn = nxn−1.
Proof.As we showed above,
(x+ h)n = xn + nxn−1h+ (stuff with at least two hs in it)
So(x+ h)n − xn
h=
nxn−1h+ (stuff with at least two hs in it)h
= nxn−1 + (stuff with at least one h in it)
and this tends to nxn−1 as h → 0.
Proving the Power RuleTheorem (The Power Rule)
Let n be a posi ve whole number. Thenddx
xn = nxn−1.
Proof.As we showed above,
(x+ h)n = xn + nxn−1h+ (stuff with at least two hs in it)
So(x+ h)n − xn
h=
nxn−1h+ (stuff with at least two hs in it)h
= nxn−1 + (stuff with at least one h in it)
and this tends to nxn−1 as h → 0.
The Power Rule for constants?Theorem
Let c be a constant. Thenddx
c = 0
..
likeddx
x0 = 0x−1
.
Proof.Let f(x) = c. Then
f(x+ h)− f(x)h
=c− ch
= 0
So f′(x) = limh→0
0 = 0.
The Power Rule for constants?Theorem
Let c be a constant. Thenddx
c = 0..
likeddx
x0 = 0x−1
.
Proof.Let f(x) = c. Then
f(x+ h)− f(x)h
=c− ch
= 0
So f′(x) = limh→0
0 = 0.
The Power Rule for constants?Theorem
Let c be a constant. Thenddx
c = 0..
likeddx
x0 = 0x−1
.
Proof.Let f(x) = c. Then
f(x+ h)− f(x)h
=c− ch
= 0
So f′(x) = limh→0
0 = 0.
..
Calculus
Recall the Limit Laws
FactSuppose lim
x→af(x) = L and lim
x→ag(x) = M and c is a constant. Then
1. limx→a
[f(x) + g(x)] = L+M
2. limx→a
[f(x)− g(x)] = L−M
3. limx→a
[cf(x)] = cL
4. . . .
Adding functionsTheorem (The Sum Rule)
Let f and g be func ons and define
(f+ g)(x) = f(x) + g(x)
Then if f and g are differen able at x, then so is f+ g and
(f+ g)′(x) = f′(x) + g′(x).
Succinctly, (f+ g)′ = f′ + g′.
Proof of the Sum RuleProof.Follow your nose:
(f+ g)′(x) = limh→0
(f+ g)(x+ h)− (f+ g)(x)h
= limh→0
f(x+ h) + g(x+ h)− [f(x) + g(x)]h
= limh→0
f(x+ h)− f(x)h
+ limh→0
g(x+ h)− g(x)h
= f′(x) + g′(x)
Proof of the Sum RuleProof.Follow your nose:
(f+ g)′(x) = limh→0
(f+ g)(x+ h)− (f+ g)(x)h
= limh→0
f(x+ h) + g(x+ h)− [f(x) + g(x)]h
= limh→0
f(x+ h)− f(x)h
+ limh→0
g(x+ h)− g(x)h
= f′(x) + g′(x)
Proof of the Sum RuleProof.Follow your nose:
(f+ g)′(x) = limh→0
(f+ g)(x+ h)− (f+ g)(x)h
= limh→0
f(x+ h) + g(x+ h)− [f(x) + g(x)]h
= limh→0
f(x+ h)− f(x)h
+ limh→0
g(x+ h)− g(x)h
= f′(x) + g′(x)
Proof of the Sum RuleProof.Follow your nose:
(f+ g)′(x) = limh→0
(f+ g)(x+ h)− (f+ g)(x)h
= limh→0
f(x+ h) + g(x+ h)− [f(x) + g(x)]h
= limh→0
f(x+ h)− f(x)h
+ limh→0
g(x+ h)− g(x)h
= f′(x) + g′(x)
Scaling functionsTheorem (The Constant Mul ple Rule)
Let f be a func on and c a constant. Define
(cf)(x) = cf(x)
Then if f is differen able at x, so is cf and
(cf)′(x) = c · f′(x)
Succinctly, (cf)′ = cf′.
Proof of Constant Multiple Rule
Proof.Again, follow your nose.
(cf)′(x) = limh→0
(cf)(x+ h)− (cf)(x)h
= limh→0
cf(x+ h)− cf(x)h
= c limh→0
f(x+ h)− f(x)h
= c · f′(x)
Proof of Constant Multiple Rule
Proof.Again, follow your nose.
(cf)′(x) = limh→0
(cf)(x+ h)− (cf)(x)h
= limh→0
cf(x+ h)− cf(x)h
= c limh→0
f(x+ h)− f(x)h
= c · f′(x)
Proof of Constant Multiple Rule
Proof.Again, follow your nose.
(cf)′(x) = limh→0
(cf)(x+ h)− (cf)(x)h
= limh→0
cf(x+ h)− cf(x)h
= c limh→0
f(x+ h)− f(x)h
= c · f′(x)
Proof of Constant Multiple Rule
Proof.Again, follow your nose.
(cf)′(x) = limh→0
(cf)(x+ h)− (cf)(x)h
= limh→0
cf(x+ h)− cf(x)h
= c limh→0
f(x+ h)− f(x)h
= c · f′(x)
Derivatives of polynomialsExample
Findddx
(2x3 + x4 − 17x12 + 37
)
Solu on
ddx
(2x3 + x4 − 17x12 + 37
)=
ddx
(2x3
)+
ddx
x4 +ddx
(−17x12
)+
ddx
(37)
= 2ddx
x3 +ddx
x4 − 17ddx
x12 + 0
= 2 · 3x2 + 4x3 − 17 · 12x11
= 6x2 + 4x3 − 204x11
Derivatives of polynomialsExample
Findddx
(2x3 + x4 − 17x12 + 37
)Solu on
ddx
(2x3 + x4 − 17x12 + 37
)=
ddx
(2x3
)+
ddx
x4 +ddx
(−17x12
)+
ddx
(37)
= 2ddx
x3 +ddx
x4 − 17ddx
x12 + 0
= 2 · 3x2 + 4x3 − 17 · 12x11
= 6x2 + 4x3 − 204x11
Derivatives of polynomialsExample
Findddx
(2x3 + x4 − 17x12 + 37
)Solu on
ddx
(2x3 + x4 − 17x12 + 37
)=
ddx
(2x3
)+
ddx
x4 +ddx
(−17x12
)+
ddx
(37)
= 2ddx
x3 +ddx
x4 − 17ddx
x12 + 0
= 2 · 3x2 + 4x3 − 17 · 12x11
= 6x2 + 4x3 − 204x11
Derivatives of polynomialsExample
Findddx
(2x3 + x4 − 17x12 + 37
)Solu on
ddx
(2x3 + x4 − 17x12 + 37
)=
ddx
(2x3
)+
ddx
x4 +ddx
(−17x12
)+
ddx
(37)
= 2ddx
x3 +ddx
x4 − 17ddx
x12 + 0
= 2 · 3x2 + 4x3 − 17 · 12x11
= 6x2 + 4x3 − 204x11
Derivatives of polynomialsExample
Findddx
(2x3 + x4 − 17x12 + 37
)Solu on
ddx
(2x3 + x4 − 17x12 + 37
)=
ddx
(2x3
)+
ddx
x4 +ddx
(−17x12
)+
ddx
(37)
= 2ddx
x3 +ddx
x4 − 17ddx
x12 + 0
= 2 · 3x2 + 4x3 − 17 · 12x11
= 6x2 + 4x3 − 204x11
OutlineDeriva ves so far
Deriva ves of power func ons by handThe Power Rule
Deriva ves of polynomialsThe Power Rule for whole number powersThe Power Rule for constantsThe Sum RuleThe Constant Mul ple Rule
Deriva ves of sine and cosine
Derivatives of Sine and CosineFactddx
sin x = ???
Proof.From the defini on:
ddx
sin x = limh→0
sin(x+ h)− sin xh
= limh→0
( sin x cos h+ cos x sin h)− sin xh
= sin x · limh→0
cos h− 1h
+ cos x · limh→0
sin hh
= sin x · 0+ cos x · 1 = cos x
Derivatives of Sine and CosineFactddx
sin x = ???
Proof.From the defini on:
ddx
sin x = limh→0
sin(x+ h)− sin xh
= limh→0
( sin x cos h+ cos x sin h)− sin xh
= sin x · limh→0
cos h− 1h
+ cos x · limh→0
sin hh
= sin x · 0+ cos x · 1 = cos x
Angle addition formulasSee Appendix A
..sin(A+ B) = sinA cos B+ cosA sin Bcos(A+ B) = cosA cos B− sinA sin B
Derivatives of Sine and CosineFactddx
sin x = ???
Proof.From the defini on:
ddx
sin x = limh→0
sin(x+ h)− sin xh
= limh→0
( sin x cos h+ cos x sin h)− sin xh
= sin x · limh→0
cos h− 1h
+ cos x · limh→0
sin hh
= sin x · 0+ cos x · 1 = cos x
Derivatives of Sine and CosineFactddx
sin x = ???
Proof.From the defini on:
ddx
sin x = limh→0
sin(x+ h)− sin xh
= limh→0
( sin x cos h+ cos x sin h)− sin xh
= sin x · limh→0
cos h− 1h
+ cos x · limh→0
sin hh
= sin x · 0+ cos x · 1 = cos x
Two important trigonometriclimitsSee Section 1.4
.. θ.
sin θ
.1− cos θ
.
θ
.−1
.1
..limθ→0
sin θθ
= 1
limθ→0
cos θ − 1θ
= 0
Derivatives of Sine and CosineFactddx
sin x = ???
Proof.From the defini on:
ddx
sin x = limh→0
sin(x+ h)− sin xh
= limh→0
( sin x cos h+ cos x sin h)− sin xh
= sin x · limh→0
cos h− 1h
+ cos x · limh→0
sin hh
= sin x · 0+ cos x · 1
= cos x
Derivatives of Sine and CosineFactddx
sin x = ???
Proof.From the defini on:
ddx
sin x = limh→0
sin(x+ h)− sin xh
= limh→0
( sin x cos h+ cos x sin h)− sin xh
= sin x · limh→0
cos h− 1h
+ cos x · limh→0
sin hh
= sin x · 0+ cos x · 1
= cos x
Derivatives of Sine and CosineFactddx
sin x = cos x
Proof.From the defini on:
ddx
sin x = limh→0
sin(x+ h)− sin xh
= limh→0
( sin x cos h+ cos x sin h)− sin xh
= sin x · limh→0
cos h− 1h
+ cos x · limh→0
sin hh
= sin x · 0+ cos x · 1 = cos x
Illustration of Sine and Cosine
.. x.
y
.π
.−π
2
.0.
π2
.π
.
sin x
.cos x
I f(x) = sin x has horizontal tangents where f′ = cos(x) is zero.I what happens at the horizontal tangents of cos?
Illustration of Sine and Cosine
.. x.
y
.π
.−π
2
.0.
π2
.π
.
sin x
.cos x
I f(x) = sin x has horizontal tangents where f′ = cos(x) is zero.
I what happens at the horizontal tangents of cos?
Illustration of Sine and Cosine
.. x.
y
.π
.−π
2
.0.
π2
.π
.
sin x
.cos x
I f(x) = sin x has horizontal tangents where f′ = cos(x) is zero.I what happens at the horizontal tangents of cos?
Derivative of CosineFactddx
cos x = − sin x
Proof.We already did the first. The second is similar (muta s mutandis):
ddx
cos x = limh→0
cos(x+ h)− cos xh
= limh→0
(cos x cos h− sin x sin h)− cos xh
= cos x · limh→0
cos h− 1h
− sin x · limh→0
sin hh
= cos x · 0− sin x · 1 = − sin x
Derivative of CosineFactddx
cos x = − sin x
Proof.We already did the first. The second is similar (muta s mutandis):
ddx
cos x = limh→0
cos(x+ h)− cos xh
= limh→0
(cos x cos h− sin x sin h)− cos xh
= cos x · limh→0
cos h− 1h
− sin x · limh→0
sin hh
= cos x · 0− sin x · 1 = − sin x
Derivative of CosineFactddx
cos x = − sin x
Proof.We already did the first. The second is similar (muta s mutandis):
ddx
cos x = limh→0
cos(x+ h)− cos xh
= limh→0
(cos x cos h− sin x sin h)− cos xh
= cos x · limh→0
cos h− 1h
− sin x · limh→0
sin hh
= cos x · 0− sin x · 1 = − sin x
Derivative of CosineFactddx
cos x = − sin x
Proof.We already did the first. The second is similar (muta s mutandis):
ddx
cos x = limh→0
cos(x+ h)− cos xh
= limh→0
(cos x cos h− sin x sin h)− cos xh
= cos x · limh→0
cos h− 1h
− sin x · limh→0
sin hh
= cos x · 0− sin x · 1 = − sin x
Derivative of CosineFactddx
cos x = − sin x
Proof.We already did the first. The second is similar (muta s mutandis):
ddx
cos x = limh→0
cos(x+ h)− cos xh
= limh→0
(cos x cos h− sin x sin h)− cos xh
= cos x · limh→0
cos h− 1h
− sin x · limh→0
sin hh
= cos x · 0− sin x · 1 = − sin x
SummaryWhat have we learned today?
I The Power Rule
I The deriva ve of a sum is the sum of the deriva vesI The deriva ve of a constant mul ple of a func on is thatconstant mul ple of the deriva ve
I The deriva ve of sine is cosineI The deriva ve of cosine is the opposite of sine.
SummaryWhat have we learned today?
I The Power RuleI The deriva ve of a sum is the sum of the deriva vesI The deriva ve of a constant mul ple of a func on is thatconstant mul ple of the deriva ve
I The deriva ve of sine is cosineI The deriva ve of cosine is the opposite of sine.
SummaryWhat have we learned today?
I The Power RuleI The deriva ve of a sum is the sum of the deriva vesI The deriva ve of a constant mul ple of a func on is thatconstant mul ple of the deriva ve
I The deriva ve of sine is cosineI The deriva ve of cosine is the opposite of sine.
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