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Chapter 2
Derivativesup down return end
2.6 Implicit differentiation
1) Explicit function: The function which can be described by expressing one variable explicitly in terms of another variable (other variables) are generally called explicit function---for example, y=xtanx, or y=[1+x2+x3]1/2 , or in general y=f(x).
2) Implicit function: The functions which are defined implicitly by a relation between variables--x and y--are generally called implicit functions--- such as x2+y2 =4, or 7sin(xy)=x2+y3 or, in general F(x,y)=0.If y=f(x) satisfies F(x, f(x))=0 on an interval I, we say f(x) is a function defined on I implicitly by F(x,y)=0, or implicit function defined by F(x,y)=0 .
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3) Derivatives of implicit function Suppose y=f(x) is an implicit function defined by sin(xy)= x2+y3. Then sin[xf(x)]=x2+ [f(x)]3. From the equation, we can find the derivative of f(x) even though we have not gotten the expression of f(x). Fortunately it is not necessary to solve the equation for y in terms of x to find the derivative. We will use the method called implicit differentiation to find the derivative. Differentiating both sides of the equation, we obtain that [f(x)+xf ' (x)]cos[xf(x)]=2x+3[f(x)]2f ' (x). Then
)cos(3
2)cos(
)](cos[)](3[
2)]([cos)('
22 xyxy
xxyy
xxfxxf
x-xxff(x)xf
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Example (a) If x3+y3 =27, find
.dx
dy
(b) Find the equation of the tangent to the curve x3+y3 =28 at point (1,3).
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Example (a) If x3+y3 =6xy, find y'. (b) Find the equation of the tangent to the
folium of Descartes x3+y3 =6xy at point (3,3).
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Orthogonal: Two curves are called Orthogonal, if at each point of intersection their tangent lines are perpendicular. If two families of curves satisfy that every curve in one family is orthogonal to every curve in the another fa
mily, then we say the two families of curves are orthogonal trajectores of each other.
Example The equations xy=c (c0) represents a family of hyperbolas. And the The equations x2-y2=k (k0) represents another family of hyperbolas with asymptotes y=x. Then the two families of curves are trajectores of each other.
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Derivative f' (x) of differentiable function f(x) is also a function. If f' (x) is differentiable, then we have [f ' (x)] '. We will denote it by f ' ' (x), i.e., f' ' (x)=[f ' (x)] '. The new function f ' ' (x) is called the second derivative of f(x). If y=f(x), we also can use other notations:
Similarly f ' ' ' (x)=[f '' (x)] ' is called the third derivative of f(x), and
)()()(
))(
('')('' 222
2
2
2
xfDxfDdx
xfddx
xdfdxd
dxyd
yxf x
2.7 Higher derivatives
)()()(
))(
(''')(''' 333
3
2
2
3
3
xfDxfDdx
xfddx
xfddxd
dxyd
yxf xup down return end
And we can define f' ' ' ' (x)=[f ' ' ' (x)] '. From now on instead of using f' ' ' ' (x) we use f(4)(x) to represent f ' ' ' ' (x). In general, we define f(n)(x)=[f(n-1)(x)] ', which is called the nth derivative of f(x). We also like to use the following notations, if y=f(x),
Example If y=x4-3x2+6x+9, find y ', y ' ', y ' ' ', y(4).
)()()(
))(
()( 1
1)()( xfDxfD
dxxfd
dxxfd
dxd
dxyd
yxf nx
nn
n
n
n
n
nnn
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Example If f(x)= , find f(n)(x). x1
Example If f(x)=sinx, g(x)=cosx, find f(n)(x) and g(n)(x) .
Example Find y ' ' , if x4+y3 =x-y .
2.8 Related rates (omitted)
2.9 Differentials, Linear and Quadratic Approximations
Definition: Let x=x-x0, f(x) =f(x)-f(x0). If there exists a constant A(x0) which is independent of x and x
such that f(x)=A(x0) x+B(x, x0) where
B(x, x0) satisfies . Then A x is called differential of f(x) at x0. Generally A x is denoted by
df(x)|x=x0 = A(x0) x. Replacing x0 by x, the differential i
s denoted by df(x) and df(x)= A(x) x.
0x
)xB(x,lim 0
0
x
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Proof: From the definition,
Corollary: If the differential of f(x) is df(x)= A(x) x, then f(x) is differentiable and A(x)=f '(x).
).(),()(
lim)()(
lim)('0
xAt
xtBtxAxt
xftfxf
txt
Corollary: (a) If f(x)=x, then dx=df(x)=x.
(b) If f(x) is differentiable, then differential of f(x) exists and df(x)=f '(x)dx.
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Example (a) Find dy, if y=x3+5x4.
(b) Find the value of dy when x=2 and dx=0.1.
Solution:
Geometric meaning of differential of f(x), df(x)=QS
f(x)=RSx
o x
y
P
t
S
RQ
dx=x
dy
y=f(x)
As x=dx is very small, y=dy ,i.e., f(t)-f(x) f '(x) t.up down return end
Example Use differentials to find an approximate (65)1/3 .
From definition of the differential, we can easily get
If f(x) is differentiable at x=a, and x is very closed to a, then f(x) f(a)+f '(a)(x-a). The approximation is called Linear approximation or tangent line approximation of f(x) at a. And function L(x)= f(a)+f '(a)(x-a) is called the linearization of f(x) at a.
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Example Find the linearization of the function f(x)=(x+3)1/2 and approximations the numbers (3.98) 1/2 and (4.05)1/2.
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Quadratic approximation to f(x) near x=a: Suppose f(x) is a function which the second derivative f ' '(a) exists. P(x)=A+Bx+Cx2 is the parabola which satisfies P(a)=f(a), P '(a)=f '(a), and P ' '(a)=f ' '(a). As x is very closed to a, the P(x) is called Quadratic approximation to f(x) near a.
Corolary: Suppose P(x)=A+Bx+Cx2 is the Quadratic approximation to f(x) near a. Then
P(x)=f(a)+f '(a)(x-a)+ f' '(a)(x-a)2 / 2.
If P(x) is the quadratic approximation to f(x) near x=a, then as x is very closed to a, P(x) f(x).That is
f(x) f(a)+f '(a)(x-a)+ f' '(a)(x-a)1/2/2.
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Example Find the quadratic approximation to f(x)=cosx near 0.
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Example Find the quadratic approximation to f(x)=(x+3)1/2 near x=1.
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The method is to give a way to get a approximation to a root of an equation.
2.10 Newton’s method(to be omitted)
Suppose f(x) is defined on [a,b], f '(x) does not value 0. Let x0[a,b], f(a)f(b)<0.
And x1=x0- , x2=x1- . Keeping repeating the process (xn=xn-1- ), we obtain a sequence of approximations x1 , x2 ,..., xn ,...... If
, then r is the root of the equation f(x)=0.
)(xf')f(x
0
0
)(xf')f(x
1-n
1-n
)(xf')f(x
1
1
rxnn
lim
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Example Starting with x1=2, find the third approximation x3 to the root of the equation x3-2x-5=0.
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2.1 Derivatives
We defined the slope of the tangent to a curve with equation y=f(x) at the point x=a to be
hm f(a)-h)f(a
lim0
h
Generally we give the following definition:
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Definition: The derivative of a function f at a number a, denoted by f´(a), is
if this limit exists.
hf(a)-h)f(a
lim0h
Then we have:
ax
afxfaf
ax -
)(-)(lim)('
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Example Find the derivative of the function y=x2-8x+9 at a.
Geometric interpretation: The derivative of the function y=f(x) at a is the slope of tangent line to y=f(x) at (a, f(a)). The line is through (a, f(a)).So if f ´(a) exists, the equation of the tangent line to the curve y=f(x) at (a, f(a)) is y-f(a)= f ´(a) (x-a).
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Example Find the equation of the tangent line of the function y=x2-3x+5 at x=1.
In the definition if we replace a by x, then we obtain a new function f ´(x) which is deduced from f(x).
)()(lim)('
0 h
x-fhxfxf
h
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Example If f(x)=(x-1)1/2, find the derivative of f . State the domain of f´(x).
Example Find the derivative of f if
1-x f(x)= 2+x
Other notations: If y=f(x), then the other notations are that f´(x)= y´= = = =Df(x)=Dxf(x).dx
dydxdf )(xf
dxd
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The symbol D and d/dx are called differential operators.
We also use the notations:
axax dx
df
dx
dfaf
)('
Definition A function f is called differentiable at a if f´(a) exists. It is differentiable on an open interval (a,b) [or (a,+) or (- ,b) ] if it is differentiable at every number in the interval.
Example Where is the function f(x)=|x| is differentiable?
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Theorem: If f(x) is differentiable at a, then f(x) is continuous at a.( The converse is false)
(3) the points at which the curve has a vertical tangent line,
such as, f(x)=x1/3, at x=0.
(1) the points at which graph of the function f has “corners”,
such as f(x)=|x| at x=0;
(2) the points at which the function is not continuous, such as, the function, defined as f(x)=2x for x1, and 3x for x<1, at x=1;
There are several cases a function fails to be differentiable
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2.2 Differentiation
1). Theorem If f is a constant function, f(x)=c, then f´(x)= (c)´=0, i.e., =0.c
dx
d
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2). The power rule If f(x)=xn, where n is a positive integer, then f´(x)= nxn-1, xn =nxn-1.dx
d
Example If f(x)=x100, find f´(x).
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3)Theorem Suppose c is a constant and f´(x) and g´(x) exist.Then
Example If f(x)= x50 +x100, find f´(x).
(c) (f(x)-g(x))´exists and (f(x)-g(x)´=f´(x)-g´(x).
(b) (f(x)+g(x))´exists and (f(x)+g(x)´=f´(x)+g´(x);
(a) (cf(x))´ exists and (cf(x))´=cf´(x);
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4) Product rule Suppose f´(x) and g´(x) exist. Then
f(x)g(x) is differentiable and
[f(x)g(x)]´= f´(x) g(x)+f(x)g´(x) .
Example If f(x)= (2x5)(3x10), find f´(x).
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4) Quotient rule Suppose f´(x) and g´(x) exist and g(x)0, then f(x)/g(x) is differentiable and
[f(x)/g(x)]´= [f´(x) g(x)-f(x)g´(x)]/[g(x)]2.
Example If f(x)= , find f´(x).x2+2x-5
x3-6
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2). The power rule (general version) If f(x)=xn, where n is any real number, then f´(x)= nxn-1,
,i.e., xn =nxn-1.dx
d
Example If f(x)=x, find f´(x).
If g(x)= x1/2, g´(x)=?
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Example Differentiate the function f(t)=(1-t)t1/3.
Table of differentiation formulas (in paper 119)
2.3 Rate of change in the Economics
Suppose C(x) is the total cost that a company
Cx
=C(x2)-C(x1) x2-x1
= C(x1+x)-C(x1) x
average of change of the cost is
the additional cost is C= C(x2)-C(x1), and the
number of items produced increased from x1 to x2,
The function C is called a cost function. If the
incurs in producing x units of certain commodity.
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The limit of this quantity as x0, is called the marginal cost by economist.
Marginal cost=dx
dC
x
Climx
Taking x=1 and n large (so that x is small compared to n),we have C'(n) C(n+1)-C(n). Thus the marginal cost of producing n is approximately equal to the cost of producing one more unit [the (n+1)st unit].
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2.4 Derivatives of trigonometric functions
(1) Theorem 0sinlim0
Proof: suppose OP=1 and (0, /2). So we will show
.0sinlim0
A
y
x
-1
C
B
o
D
Notice that 0<|BC|<arcAB
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(2) Corollary .1coslim0
(3) Theorem .1sin
lim0
x
-1
C
B
o
Dy
A
Proof: Notice that Area of OAB<Area of sector OAB<Area of OAD.
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(4) Corollary .01cos
lim0
.4sin7sin
lim0 x
xx
Example Find
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(5) Theorem .cossin xxdxd
Example Differentiate y=xsinx.
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(6) Theorem .sincos xxdxd
Example Differentiate y=tanx.
Corollary (tanx)'=sec2x
Example Differentiate y=cotx.
Corollary (cotx)'= - csc2x
Example Differentiate f(x)= .cos2
cotsintan2 xx
xxx
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2.5 Chain rule
The chain rule If the derivative g'(x) and derivative f '(u), with respect to u, exist, then the composite function f(g(x)) is differentiable, and [f(g(x))] ' =f '(g(x))g '(x).
Let u=g(x+x)-g(x) y=f(u+u)-f(u)
.limlim00 x
uuy
xy
dxdy
xx
Proof:
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Case 1: du/dx0, then u0
.limlimlim000 x
uuy
xy
dxdy
xxx
Case 2: du/dx=0. there are two cases:
.0limlimlim000
xu
uy
xy
dxdy
xxx
.00
limlim00
xxy
dxdy
xx
(a) u 0,
(b) u= 0,
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Example Find F '(x) if F(x)=(1+x2)3/4.
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