Lesson 8: Basic Differentation Rules (handout)

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NYUMathematics

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Sec on 2.3Basic Differenta on Rules

V63.0121.001: Calculus IProfessor Ma hew Leingang

New York University

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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

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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.

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Notes

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. Sec on 2.3: Basic Differenta on Rules. V63.0121.001: Calculus I .

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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.

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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.

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NotationNewtonian nota on Leibnizian nota on

f′(x) y′(x) y′dydx

ddx

f(x)dfdx

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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

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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

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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.

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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

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The squaring function and its derivatives

.. x.

y

. f.

f′

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f′′ I f increasing =⇒ f′ ≥ 0I f decreasing =⇒ f′ ≤ 0I horizontal tangent at 0=⇒ f′(0) = 0

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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.

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The cubing function and its derivatives

.. x.

y

.f

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f′

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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)

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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

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The square root and its derivatives

.. x.

y

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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

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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

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The cube root and its derivative

.. x.

y

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f

. f′

I Here limx→0

f′(x) = ∞ and fis not differen able at 0

I No ce alsolim

x→±∞f′(x) = 0

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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

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x 7→ x2/3 and its derivative

.. x.

y

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f

. f′

I f is not differen able at 0and lim

x→0±f′(x) = ±∞

I No ce alsolim

x→±∞f′(x) = 0

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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

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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.

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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

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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

. . . . . .

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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.

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The Power Rule for constants?Theorem

Let c be a constant. Thenddx

c = 0..

likeddx

x0 = 0x−1

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Proof.Let f(x) = c. Then

f(x+ h)− f(x)h

=c− ch

= 0

So f′(x) = limh→0

0 = 0.

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Calculus

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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. . . .

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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′.

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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)

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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′.

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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)

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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

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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

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Angle addition formulasSee Appendix A

..

sin(A+ B) = sinA cos B+ cosA sin Bcos(A+ B) = cosA cos B− sinA sin B

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Two important trigonometriclimitsSee Section 1.4

.. θ.

sin θ

.1− cos θ

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θ

.−1

.1

..limθ→0

sin θθ

= 1

limθ→0

cos θ − 1θ

= 0

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Illustration of Sine and Cosine

.. x.

y

.π

.−π

2

.0.

π2

.π

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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?

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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

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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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Lesson 8: Basic Differentation Rules (handout)

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