ma4001 engineering mathematics 1 lecture 15 mean value ...ma4001 engineering mathematics 1 lecture...

Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation

MA4001 Engineering Mathematics 1Lecture 15

Mean Value TheoremIncreasing and Decreasing Functions

Higher Order DerivativesImplicit Differentiation

Dr. Sarah Mitchell

Autumn 2014


Rolle’s Theorem

Theorem

If:

f (x) is continuous in [a,b];

f (x) is differentiable in (a,b);

f (a) = f (b)

then, there exists a point c such that f ′(c) = 0.


Proof.

f (a) = f (b) could mean that f (x) is constant in [a,b], in whichcase f ′ = 0 everywhere in [a,b].

Otherwise, if f (x) is not constant, by the max-min theorem, f (x)achieves a maximum and minimum in [a,b].

At least one maximum or minimum must be at an interior pointwhich we can call c.

Thus by the previous theorem, f ′(c) = 0.


Mean Value Theorem

Theorem

If f (x) is

continuous in [a,b];

differentiable in (a,b);

then there exists a point c ∈ (a,b) such that

f ′(c) =f (b) − f (a)

b − a


Note that here:

f ′(c) is the slope of the tangent at c.

f (b) − f (a)b − a

is the slope of the secant joining (a, f (a)) and

(b, f (b))

The equation of the secant is y − f (a) =f (b) − f (a)

b − a(x − a).


Proof of Mean Value Theorem

Proof.

Let g(x) = f (x) −[

f (a) +f (b) − f (a)

b − a(x − a)

]

.

Then g(x) satisfies the conditions for Rolle’s theorem:

g(x) is continuous in [a,b] and differentiable in (a,b);

g(a) = g(b) = 0.

Thus there is a point c such that g ′(c) = 0.

g ′(c) = 0 = f ′(c) −f (b) − f (a)

b − a.

Thus f ′(c) =f (b) − f (a)

b − a


Example

Prove that sin x < x for all x > 0.


Proof.

The result is obvious for x > 1 as sin x 6 1 < x .

For 0 < x < 1, consider f (t) = sin t on [0, x ].

By the mean value theorem there exists a point c such that0 < c < x and

f ′(c) =f (x) − f (0)

x − 0=

f (x)x

=sin x

x

That is,

1 > cos c =sin x

xas 0 < c < 1

Thus sin x < x .


Increasing functions

Definition

Suppose f (x) is defined on an interval I.

Then if, for all x2 > x1 ∈ I, f (x2) > f (x1), f is said to beincreasing on I.


Decreasing functions

Definition


Then if, for all x2 > x1 ∈ I, f (x2) < f (x1), f is said to bedecreasing on I.


Non-decreasing functions

Definition


Then if, for all x2 > x1 ∈ I, f (x2) > f (x1), f is said to benon-decreasing on I.


Non-increasing functions

Definition


Then if, for all x2 > x1 ∈ I, f (x2) 6 f (x1), f is said to benon-increasing on I.


Theorem

If for all x ∈ (a,b):

f ′(x) > 0, then f is increasing in (a,b);

f ′(x) < 0, then f is decreasing in (a,b);

f ′(x) > 0, then f is non-decreasing in (a,b);

f ′(x) 6 0, then f is non-increasing in (a,b).


Remark

f ′(x) > 0 means that tangent lines have positive slopes.

f ′(x) < 0 means that tangent lines have negative slopes.


Proof for f ′(x) > 0 case

For any x1, x2, such that a < x1 < x2 < b, apply the mean valuetheorem:

There exists a point c ∈ (x1, x2), such that

f (x2) − f (x1)

x2 − x1= f ′(c) > 0

Since x2 − x1 > 0, f (x2) − f (x1) must also be > 0.

Thus f (x) is increasing.

The other 3 cases can be proved analogously.


Example: On which intervals is f (x) = x3 − 12x + 1increasing or decreasing?

f ′(x) = 3x2 − 12 = 2(x + 2)(x − 2)

f ′(x) > 0, i.e., f (x) is increasing if x > 2 or x < −2 .

f ′(x) < 0, i.e., f (x) is decreasing if −2 < x < 2 .


Higher order derivatives

The derivative of y ′ = f ′(x) is called the second derivative of f :

f ′′(x) = (f ′(x)) ′

denoted in various ways as

y ′′ = f ′′(x) =d2ydx2 =

ddx

ddx

f (x) =d2

dx2 f (x)

Similarly the n-th derivative of f is

f (n)(x) =(

. . .

(

(

f ′)

′

)

′

. . .

)

′

(x)

i.e., f (x) differentiated n times.

Note that the following notations can also be used, in particularfor higher derivatives:

f (0)(x) = f (x), f (1)(x) = f ′(x), f (2)(x) = f ′′(x), f (3)(x) = f ′′′(x), . . .


Example: nth degree polynomial p(x) =xn + an−1xn−1 + an−2xn−2 + · · ·+ a3x3 + a2x2 + a1x + a0

p ′(x) =nxn−1+(n−1)an−1xn−2+(n−2)an−2xn−3+· · ·+3a3x2+2a2x+a1

p ′′(x) = n(n − 1)xn−2 + (n − 1)(n − 2)an−1xn−3 +

(n − 2)(n − 3)an−2xn−4 + · · · + 6a3x + 2a2

p ′′′(x) = n(n − 1)(n − 2)xn−3 + (n− 1)(n− 2)(n− 3)an−1xn−4 +

(n − 2)(n − 3)(n − 4)an−2xn−5 + · · · + 6a3...Differentiating n times we obtainp(n) = n(n − 1)(n − 2)(n − 3) . . . 3 · 2 · 1 = n!


Example: y =1x

y = x−1

y ′ = (−1)x−2

y ′′ = (−1)(−2)x−3

y ′′′ = (−1)(−2)(−3)x−4 ...

y(n) = (−1)(−2)(−3) . . . (−n)x−(n+1) =(−1)nn!

xn+1


Example: differential equation

If y = A cos(kx) + B sin(kx), then

y ′ = −kA sin(kx) + kB cos(kx)

y ′′ = −k2A cos(kx) − k2B sin(kx)

Therefore y satisfies the differential equation y ′′ + k2y = 0.

This is the differential equation of simple harmonic motion.

For example, it describes the motion of a mass suspended froma fixed base by a spring (with no damping).


Implicit differentiation

A function may be defined:

explicitly: y = f (x)e.g., y = x3, y =

√x , or

implicitly: i.e., through an equation F (x , y) = 0.e.g., x2 + y2 = 22.


Often we cannot solve F (x , y) = 0 to obtain an explicitrepresentation for y , but the derivative y ′ may still be defined.

We differentiate F (x , y) = 0 with respect to x , regarding y as afunction of x and using the chain rule.


Example: Finddydx

where y sin x = x3 + cos y .

We differentiate both sides of the equation with respect to x :

ddx

(y sin x) =ddx

(x3) +ddx

(cos y)

⇒ y cos x +dydx

sin x = 3x2 − sin ydydx

by the product rule by the chain rule

⇒dydx

(sin x + sin y) = 3x2 − y cos x

⇒dydx

=3x2 − y cos xsin x + sin y

ma4001 engineering mathematics 1 lecture 15 mean value ...ma4001 engineering mathematics 1 lecture...

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