Lesson 8: Basic Differentation Rules (slides)

Sec on 2.3
Basic Diﬀerenta on Rules
V63.0121.011: Calculus I
Professor Ma hew Leingang
New York University

. NYUMathematics

Announcements

Quiz 1 this week on
1.1–1.4
Quiz 2 March 3/4 on 1.5,
1.6, 2.1, 2.2, 2.3
Midterm Monday March
7 in class

Objectives
Understand and use
these diﬀeren a on
rules:
the deriva ve of a
constant func on (zero);
the Constant Mul ple
Rule;
the Sum Rule;
the Diﬀerence Rule;
the deriva ves of sine
and cosine.

Recall: the derivative

Deﬁni on
Let f be a func on and a a point in the domain of f. If the limit
f(a + h) − f(a) f(x) − f(a)
f′ (a) = lim = lim
h→0 h x→a x−a
exists, the func on is said to be diﬀeren able at a and f′ (a) is the
deriva ve of f at a.

The deriva ve …
…measures the slope of the line through (a, f(a)) tangent to
the curve y = f(x);
…represents the instantaneous rate of change of f at a
…produces the best possible linear approxima on to f near a.

Notation
Newtonian nota on Leibnizian nota on
dy d df
f′ (x) y′ (x) y′ f(x)
dx dx dx

Link between the notations
f(x + ∆x) − f(x) ∆y dy
f′ (x) = lim = lim =
∆x→0 ∆x ∆x→0 ∆x dx
dy
Leibniz thought of as a quo ent of “infinitesimals”
dx
dy
We think of as represen ng a limit of (finite) difference
dx
quo ents, not as an actual frac on itself.
The nota on suggests things which are true even though they
don’t follow from the nota on per se

Outline
Deriva ves so far
Deriva ves of power func ons by hand
The Power Rule
Deriva ves of polynomials
The Power Rule for whole number powers
The Power Rule for constants
The Sum Rule
The Constant Mul ple Rule
Deriva ves of sine and cosine

Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the deﬁni on of deriva ve to ﬁnd f′ (x).

Example

Solu on

f(x + h) − f(x)
f′ (x) = lim
h→0 h

Example

Solu on

′ f(x + h) − f(x) (x + h)2 − x2
f (x) = lim = lim
h→0 h h→0 h

Example

Solu on

′ f(x + h) − f(x) (x + h)2 − x2
f (x) = lim = lim
h→0 h h→0 h
+ 2xh + h −
2
x2
x2

= lim
h→0 h

Example

Solu on

′ f(x + h) − f(x) (x + h)2 − x2
f (x) = lim = lim
h→0 h h→0 h
+ 2xh + h −
x2
2
x2
2x + h2
h
¡
= lim = lim
h→0 h h→0 h

Example

Solu on

′ f(x + h) − f(x) (x + h)2 − x2
f (x) = lim = lim
h→0 h h→0 h
+ 2xh + h −
x2
2
x2
2x + h2
h
¡
= lim = lim
h→0 h h→0 h

= lim (2x + h) = 2x.
h→0

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!

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! Leibnizian
nota on:
d2 y d2 d2 f
f(x)
dx2 dx2 dx2

The squaring function and its derivatives

y

. x


y

. f x


y
f′
f increasing =⇒ f′ ≥ 0
f decreasing =⇒ f′ ≤ 0
. f x horizontal tangent at 0
=⇒ f′ (0) = 0


y
f′
f′′ f increasing =⇒ f′ ≥ 0
f decreasing =⇒ f′ ≤ 0
. f x horizontal tangent at 0
=⇒ f′ (0) = 0

Derivative of the cubing function
Example

Example

Solu on

′ f(x + h) − f(x) (x + h)3 − x3
f (x) = lim = lim
h→0 h h→0 h

Example

Solu on

′ f(x + h) − f(x) (x + h)3 − x3
f (x) = lim = lim
h→0 h h→0 h
x3

+ 3x2 h + 3xh2 + h3 − 3
x
= lim
h→0 h

Example

Solu on

′ f(x + h) − f(x) (x + h)3 − x3
f (x) = lim = lim
h→0 h h→0 h
1 2
!
¡ !
¡
x3

+
2 2
3x h + 3xh + h 3
− 3
x 3x2 +
h
¡
3xh2 + ¡
h3
= lim = lim
h→0 h h→0 h

Example

Solu on

′ f(x + h) − f(x) (x + h)3 − x3
f (x) = lim = lim
h→0 h h→0 h
1 2
!
¡ !
¡
x3
2 2
3x h + 3xh + h
+
3
− 3
x 3x2 +
h
¡
3xh2 + ¡
h3
= lim = lim
h→0
( 2 h ) h→0 h

= lim 3x + 3xh + h2 = 3x2 .
h→0

The cubing function and its derivatives

y

. x


y

f
. x

No ce that f is increasing,
y and f′ 0 except
f′ f′ (0) = 0

f
. x

y and f′ 0 except
f′′ f′ f′ (0) = 0

f
. x

y and f′ 0 except
f′′ f′ f′ (0) = 0
No ce also that the
f tangent line to the graph
. x of f at (0, 0) crosses the
graph (contrary to a
popular “deﬁni on” of
the tangent line)

Derivative of the square root
Example
√
Suppose f(x) = x = x1/2 . Fnd f′ (x) with the deﬁni on.

Example
√

Solu on
√ √
′ f(x + h) − f(x) x+h− x
f (x) = lim = lim
h→0 h h→0 h

Example
√

Solu on
√ √
′ f(x + h) − f(x) x+h− x
f (x) = lim = lim
h→0
√ h h→0 h
√ √ √
x+h− x x+h+ x
= lim ·√ √
h→0 h x+h+ x

Example
√

Solu on
√ √
′ f(x + h) − f(x) x+h− x
f (x) = lim = lim
h→0
√ h h→0 h
√ √ √
x+h− x x+h+ x
= lim ·√ √
h→0 h x+h+ x
(x + h) − x
¡ ¡
= lim (√ √ )
h→0 h x+h+ x

Example
√

Solu on
√ √
′ f(x + h) − f(x) x+h− x
f (x) = lim = lim
h→0
√ h h→0 h
√ √ √
x+h− x x+h+ x
= lim ·√ √
h→0 h x+h+ x
(x + h) − x
¡ ¡ h

= lim (√ √ ) = lim (√ √ )
h→0 h x+h+ x h x+h+ x
h→0

Example
√

Solu on
√ √
′ f(x + h) − f(x) x+h− x
f (x) = lim = lim
h→0
√ h h→0 h
√ √ √
x+h− x x+h+ x
= lim ·√ √
h→0 h x+h+ x
(x + h) − x
¡ ¡ h
1
= lim (√ √ ) = lim (√ √ )= √
h→0 h x+h+ x h x+h+ x
h→0 2 x

The square root and its derivatives

y

. x


y

f
. x


y

f Here lim+ f′ (x) = ∞ and f
x→0
f′ is not diﬀeren able at 0
. x


y

f Here lim+ f′ (x) = ∞ and f
x→0
f′ is not diﬀeren able at 0
. x
No ce also lim f′ (x) = 0
x→∞

Derivative of the cube root
Example
√
Suppose f(x) = 3
x = x1/3 . Find f′ (x).

Example
√
Suppose f(x) = 3
x = x1/3 . Find f′ (x).

Solu on
f(x + h) − f(x) (x + h)1/3 − x1/3
f′ (x) = lim = lim
h→0 h h→0 h

Example
√
Suppose f(x) = 3
x = x1/3 . Find f′ (x).

Solu on
f(x + h) − f(x) (x + h)1/3 − x1/3
h→0 h h→0 h
(x + h) − x
1/3 1/3
(x + h) + (x + h)1/3 x1/3 + x2/3
2/3
= lim ·
h→0 h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3

Example
√
Suppose f(x) = 3
x = x1/3 . Find f′ (x).

Solu on
f(x + h) − f(x) (x + h)1/3 − x1/3
h→0 h h→0 h
(x + h) − x
1/3 1/3
(x + h) + (x + h)1/3 x1/3 + x2/3
2/3
= lim ·
h→0 h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
(x + h) − x
¡ ¡
= lim 2/3 + (x + h)1/3 x1/3 + x2/3 )
h→0 h ((x + h)

Example
√
Suppose f(x) = 3
x = x1/3 . Find f′ (x).

Solu on
f(x + h) − f(x) (x + h)1/3 − x1/3
h→0 h h→0 h
(x + h) − x
1/3 1/3
(x + h) + (x + h)1/3 x1/3 + x2/3
2/3
= lim ·
h→0 h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
(x + h) − x
¡ ¡
= lim 2/3 + (x + h)1/3 x1/3 + x2/3 )
h→0 h ((x + h)

h

= lim 2/3 + (x + h)1/3 x1/3 + x2/3 )
h→0 ((x + h)
h

Example
√
Suppose f(x) = 3
x = x1/3 . Find f′ (x).

Solu on
f(x + h) − f(x) (x + h)1/3 − x1/3
h→0 h h→0 h
(x + h) − x
1/3 1/3
(x + h) + (x + h)1/3 x1/3 + x2/3
2/3
= lim ·
h→0 h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
(x + h) − x
¡ ¡
= lim 2/3 + (x + h)1/3 x1/3 + x2/3 )
h→0 h ((x + h)

h
1
= lim 2/3 + (x + h)1/3 x1/3 + x2/3 )
= 2/3
h→0 ((x + h)
h 3x

The cube root and its derivative

y

. x


y

f
. x


y
Here lim f′ (x) = ∞ and f
f x→0
is not diﬀeren able at 0
f′
. x


y
Here lim f′ (x) = ∞ and f
f x→0
is not diﬀeren able at 0
f′
. x No ce also
lim f′ (x) = 0
x→±∞

One more
Example
Suppose f(x) = x2/3 . Use the deﬁni on of deriva ve to ﬁnd f′ (x).

One more
Example

Solu on
f(x + h) − f(x) (x + h)2/3 − x2/3
h→0 h h→0 h

One more
Example

Solu on
f(x + h) − f(x) (x + h)2/3 − x2/3
h→0 h h→0 h
1/3 ( )
(x + h) − x
1/3
= lim · (x + h) + x
1/3 1/3
h→0 h

One more
Example

Solu on
f(x + h) − f(x) (x + h)2/3 − x2/3
h→0 h h→0 h
1/3 ( )
(x + h) − x
1/3
= lim · (x + h) + x
1/3 1/3
h→0
( h)
1 −2/3
= 3x 2x1/3

One more
Example

Solu on
f(x + h) − f(x) (x + h)2/3 − x2/3
h→0 h h→0 h
1/3 ( )
(x + h) − x
1/3
= lim · (x + h) + x
1/3 1/3
h→0
( h)
1 −2/3
= 3x 2x1/3 = 2 x−1/3
3

x → x2/3 and its derivative

y

. x


y
f

. x


y
f f is not diﬀeren able at 0
and lim± f′ (x) = ±∞
f′ x→0
. x


y
f f is not diﬀeren able at 0
and lim± f′ (x) = ±∞
f′ x→0
. x No ce also
lim f′ (x) = 0
x→±∞

Recap
y y′
x2 2x
x3 3x2
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x

Recap
y y′
x2 2x1
x3 3x2
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x

Recap: The Tower of Power
y y′
x2 2x1 The power goes down by
x 3
3x2 one in each deriva ve
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x

y y′
x2 2x The power goes down by
x 3
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x

y y′
x2 2x The power goes down by
x 3
1 −1/2 The coeﬃcient in the
x1/2 2x deriva ve is the power of
1 −2/3
x1/3 3x
the original func on
2 −1/3
x2/3 3x

The Power Rule
There 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.

Perhaps the most famous rule in calculus
We will assume it as of today
We will prove it many ways for many diﬀerent r.

Remember your algebra
Fact
Let n be a posi ve whole number. Then

(x + h)n = xn + nxn−1 h + (stuﬀ with at least two hs in it)

Remember your algebra
Fact
Let n be a posi ve whole number. Then


Proof.
We have
∑
n
(x + h) = (x + h) · (x + h) · (x + h) · · · (x + h) =
n
ck xk hn−k
n copies k=0

Pascal’s Triangle
.
1
1 1
1 2 1
1 3 3 1 (x + h)0 = 1
1 4 6 4 1 (x + h)1 = 1x + 1h
(x + h)2 = 1x2 + 2xh + 1h2
1 5 10 10 5 1
(x + h)3 = 1x3 + 3x2 h + 3xh2 + 1h3
1 6 15 20 15 6 1 ... ...

Proving the Power Rule
d n
Let n be a posi ve whole number. Then x = nxn−1 .
dx

Proving the Power Rule
d n
Let n be a posi ve whole number. Then x = nxn−1 .
dx

Proof.
As we showed above,


(x + h)n − xn nxn−1 h + (stuﬀ with at least two hs in it)
So =
h 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
d
Let c be a constant. Then c=0
dx

d 0
Theorem like x = 0x−1
d dx
Let c be a constant. Then c = 0.
.
dx

d 0
Theorem like x = 0x−1
d dx
Let c be a constant. Then c = 0.
.
dx

Proof.
Let f(x) = c. Then

f(x + h) − f(x) c − c
= =0
h h
So f′ (x) = lim 0 = 0.
h→0

Recall the Limit Laws
Fact
Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then
x→a x→a
1. lim [f(x) + g(x)] = L + M
x→a
2. lim [f(x) − g(x)] = L − M
x→a
3. lim [cf(x)] = cL
x→a
4. . . .

Adding functions
Theorem (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 Rule
Proof.
Follow your nose:
(f + g)(x + h) − (f + g)(x)
(f + g)′ (x) = lim
h→0 h

Proof.
Follow your nose:
(f + g)(x + h) − (f + g)(x)
(f + g)′ (x) = lim
h→0 h
f(x + h) + g(x + h) − [f(x) + g(x)]
= lim
h→0 h

Proof.
Follow your nose:
(f + g)(x + h) − (f + g)(x)
(f + g)′ (x) = lim
h→0 h
f(x + h) + g(x + h) − [f(x) + g(x)]
= lim
h→0 h
f(x + h) − f(x) g(x + h) − g(x)
= lim + lim
h→0 h h→0 h

Proof.
Follow your nose:
(f + g)(x + h) − (f + g)(x)
(f + g)′ (x) = lim
h→0 h
f(x + h) + g(x + h) − [f(x) + g(x)]
= lim
h→0 h
f(x + h) − f(x) g(x + h) − g(x)
= lim + lim
h→0 h h→0 h
′ ′
= f (x) + g (x)

Scaling functions
Theorem (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 + h) − (cf)(x)
(cf)′ (x) = lim
h→0 h

Proof.
(cf)(x + h) − (cf)(x) cf(x + h) − cf(x)
(cf)′ (x) = lim = lim
h→0 h h→0 h

Proof.
(cf)(x + h) − (cf)(x) cf(x + h) − cf(x)
h→0 h h→0 h
f(x + h) − f(x)
= c lim
h→0 h

Proof.
(cf)(x + h) − (cf)(x) cf(x + h) − cf(x)
h→0 h h→0 h
f(x + h) − f(x)
= c lim = c · f′ (x)
h→0 h

Derivatives of polynomials
Example
d ( 3 )
Find 2x + x4 − 17x12 + 37
dx

Example
d ( 3 )
Find 2x + x4 − 17x12 + 37
dx
Solu on

d ( 3 ) d ( 3) d d ( ) d
2x + x4 − 17x12 + 37 = 2x + x4 + −17x12 + (37)
dx dx dx dx dx

Example
d ( 3 )
Find 2x + x4 − 17x12 + 37
dx
Solu on

d ( 3 ) d ( 3) d d ( ) d
2x + x4 − 17x12 + 37 = 2x + x4 + −17x12 + (37)
dx dx dx dx dx
d d d
= 2 x3 + x4 − 17 x12 + 0
dx dx dx

Example
d ( 3 )
Find 2x + x4 − 17x12 + 37
dx
Solu on

d ( 3 ) d ( 3) d d ( ) d
2x + x4 − 17x12 + 37 = 2x + x4 + −17x12 + (37)
dx dx dx dx dx
d d d
= 2 x3 + x4 − 17 x12 + 0
dx dx dx
= 2 · 3x + 4x − 17 · 12x11
2 3

Example
d ( 3 )
Find 2x + x4 − 17x12 + 37
dx
Solu on

d ( 3 ) d ( 3) d d ( ) d
2x + x4 − 17x12 + 37 = 2x + x4 + −17x12 + (37)
dx dx dx dx dx
d d d
= 2 x3 + x4 − 17 x12 + 0
dx dx dx
= 2 · 3x + 4x − 17 · 12x11
2 3

= 6x2 + 4x3 − 204x11

Derivatives of Sine and Cosine
Fact
d
sin x = ???
dx
Proof.
From the deﬁni on:

Fact
d
sin x = ???
dx
Proof.
d sin(x + h) − sin x
sin x = lim
dx h→0 h

Angle addition formulas
See Appendix A

sin(A + B) = . A cos B + cos A sin B
sin
cos(A + B) = cos A cos B − sin A sin B

Fact
d
sin x = ???
dx
Proof.
d sin(x + h) − sin x ( sin x cos h + cos x sin h) − sin x
sin x = lim = lim
dx h→0 h h→0 h

Fact
d
sin x = ???
dx
Proof.
sin x = lim = lim
dx h→0 h h→0 h
cos h − 1 sin h
= sin x · lim + cos x · lim
h→0 h h→0 h

Two important trigonometric
limits
See Section 1.4

sin θ
lim . =1
θ→0 θ
cos θ − 1
sin θ θ lim =0
θ θ→0 θ
.
1 − cos θ 1

Fact
d
sin x = ???
dx
Proof.
sin x = lim = lim
dx h→0 h h→0 h
cos h − 1 sin h
h→0 h h→0 h
= sin x · 0 + cos x · 1

Fact
d
sin x = cos x
dx
Proof.
sin x = lim = lim
dx h→0 h h→0 h
cos h − 1 sin h
h→0 h h→0 h
= sin x · 0 + cos x · 1 = cos x

Illustration of Sine and Cosine
y

. x
π −π 0 π π
2 2
sin x

y

. x
π −π 0 π π cos x
2 2
sin x

f(x) = sin x has horizontal tangents where f′ = cos(x) is zero.

y

. x
π −π 0 π π cos x
2 2
sin x

f(x) = sin x has horizontal tangents where f′ = cos(x) is zero.
what happens at the horizontal tangents of cos?

Derivative of Cosine
Fact
d
cos x = − sin x
dx

Fact
d
cos x = − sin x
dx
Proof.
We already did the ﬁrst. The second is similar (muta s mutandis):
d cos(x + h) − cos x
cos x = lim
dx h→0 h

Fact
d
cos x = − sin x
dx
Proof.
d cos(x + h) − cos x (cos x cos h − sin x sin h) − cos x
cos x = lim = lim
dx h→0 h h→0 h

Fact
d
cos x = − sin x
dx
Proof.
cos x = lim = lim
dx h→0 h h→0 h
cos h − 1 sin h
= cos x · lim − sin x · lim
h→0 h h→0 h

Fact
d
cos x = − sin x
dx
Proof.
cos x = lim = lim
dx h→0 h h→0 h
cos h − 1 sin h
= cos x · lim − sin x · lim
h→0 h h→0 h
= cos x · 0 − sin x · 1 = − sin x

Summary
What have we learned today?

The Power Rule

Summary

The Power Rule
The deriva ve of a sum is the sum of the deriva ves
The deriva ve of a constant mul ple of a func on is that
constant mul ple of the deriva ve

Summary

The Power Rule
The deriva ve of a sum is the sum of the deriva ves
The deriva ve of a constant mul ple of a func on is that
constant mul ple of the deriva ve
The deriva ve of sine is cosine
The deriva ve of cosine is the opposite of sine.

Lesson 8: Basic Differentation Rules (slides)

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Lesson 8: Basic Differentation Rules (slides)