ma4001 engineering mathematics 1 lecture 15 mean value ...ma4001 engineering mathematics 1 lecture...
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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
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
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.
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
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.
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
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
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
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).
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
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
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
Example
Prove that sin x < x for all x > 0.
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
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 .
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
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.
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
Decreasing 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 bedecreasing on I.
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
Non-decreasing 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 benon-decreasing on I.
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
Non-increasing functions
Definition
Suppose f (x) is defined on an interval I.
Then if, for all x2 > x1 ∈ I, f (x2) 6 f (x1), f is said to benon-increasing on I.
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
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).
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
Remark
f ′(x) > 0 means that tangent lines have positive slopes.
f ′(x) < 0 means that tangent lines have negative slopes.
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
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.
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
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 .
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
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), . . .
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
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!
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
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
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
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).
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
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.
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
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.
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Mean Value Theorem Increasing and decreasing functions Higher Order Derivatives Implicit Differentiation
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