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Chapter 2: Differentiation
Spring 2018
Department of MathematicsHong Kong Baptist University
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§2.1 Tangent Lines and Their Slopes
This section deals with the problem of finding a straight line Lthat is tangent to a curve C at a point P.
Let C be the graph of y = f (x), and P be the point (x0, y0)on C so that y0 = f (x0). Also assume that P is not anendpoint of C .
A reasonable definition of tangency can be stated in terms oflimits (of the slopes of secant lines).
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Given the curve y = f (x) and the point P = (x0, f (x0)), wechoose a different point Q on the curve C so thatQ = (x0 + h, f (x0 + h)). Note that h can be positive ornegative, depending on whether Q is to the right or left of P.
The line through P and Q is called a secant line to thecurve. The slope of the line PQ is
f (x0 + h)− f (x0)
h.
This expression is called the Newton quotient or differencequotient for f at x0.
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Nonvertical tangent line
Definition
Suppose that the function f is continuous at x = x0 and that
limh→0
f (x0 + h)− f (x0)
h= m (m is finite)
exists. Then the straight line having slope m and passing throughthe point P = (x0, f (x0)) is called the tangent line (or simply thetangent) to the graph of y = f (x) at P. An equation of thistangent line is
y = m(x − x0) + y0.
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Example 1
Find an equation of the tangent line to the curve y = x2 at thepoint (1, 1).
Solution: Note that f (x) = x2, x0 = 1 and y0 = f (x0) = 1. Bydefinition, the slope of the tangent line is
m = limh→0
f (1 + h)− f (1)
h
= limh→0
(1 + h)2 − 1
h= lim
h→0(2 + h)
= 2.
Accordingly, the equation of the tangent line is
y = m(x − x0) + y0 = 2(x − 1) + 1 = 2x − 1.
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(a) Vertical tangents or (b) no tangent line
Definition
(a) If f is continuous at P = (x0, y0), where y0 = f (x0), and ifeither
limh→0
f (x0 + h)− f (x0)
h=∞ or lim
h→0
f (x0 + h)− f (x0)
h= −∞,
then the vertical line x = x0 is tangent to the graph y = f (x) at P.
(b) If the limit of the Newton quotient fails to exist in any otherway than by being ∞ or −∞, the graph y = f (x) has no tangentline at P.
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Example 2
Find the tangent line to the curve y = x1/3 at the point (0, 0).
Solution: For this curve, the limit of the Newton quotient for f atx = 0 is
limh→0
f (0 + h)− f (0)
h= lim
h→0
h1/3
h= lim
h→0
1
h2/3=∞.
Thus by definition, the vertical line x = 0 (i.e., the y -axis) is thetangent line to the curve y = x1/3 at the point (0, 0).
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Example 3
Does the graph of y = |x | have a tangent line at x = 0?
Solution: For this curve, the Newton quotient is
f (x0 + h)− f (x0)
h=|0 + h| − |0|
h= sgn(h).
Now since sgn(h) has different right and left limits at 0, theNewton quotient has no limit at h→ 0. This implies that y = |x |has no tangent line at (0, 0).
Remark: The curve suddenly changes the direction and is notsmooth at x = 0.
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The slope of a curve
Definition
The slope of a curve C at a point P is the slope of the tangentline to C at P if such a tangent line exists. In particular, the slopeof the graph of y = f (x) at the point x0 is
limh→0
f (x0 + h)− f (x0)
h.
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Example 4
Find the slope of the curve y = x/(2x + 3) at the point x = 3.
Solution: The slope of the curve at x = 3 is
m = limh→0
f (3 + h)− f (3)
h
= limh→0
3+h9+2h −
13
h
= limh→0
1
27 + 6h
=1
27.
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§2.2 The Derivative
Definition
The derivative of a function f is another function f ′ defined by
f ′(x) = limh→0
f (x + h)− f (x)
h
at all points x for which the limit exists (i.e., is a finite realnumber). If f ′(x) exists, we say that f is differentiable at x .
Remark: The derivative f ′ is read as “f prime”. The derivative off is essentially the slope function of f . The process of calculatingthe derivative f ′ of a given function f is called differentiation.
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Differentiability on Intervals
A function f is differentiable on an open interval (a, b) iff ′(x) exists for all x in (a, b).
A function f is differentiable on a closed interval [a, b] if f ′(x)exists for all x in (a, b) and f ′+(a) and f ′−(b) both exist, where
f ′+(a) = limh→0+
f (a + h)− f (a)
h,
and
f ′−(b) = limh→0−
f (b + h)− f (b)
h.
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Example 5: (The derivative of a linear function)
Show that if f (x) = ax + b, then f ′(x) = a.
Solution: By definition, we have
f ′(x) = limh→0
f (x + h)− f (x)
h
= limh→0
a(x + h) + b − (ax + b)
h
= limh→0
ah
h= a.
A special case: If g(x) = c (constant), then g ′(x) = 0.
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Theorem (Power Rule)
If f (x) = xn, then f ′(x) = nxn−1.
Proof (when n is a positive integer): To prove the theorem, weneed the following factorization formula:
an − bn = (a− b)(an−1 + an−2b + · · ·+ abn−2 + bn−1).
Then
f ′(x) = limh→0
(x + h)n − xn
h
= limh→0
[(x + h)n−1 + (x + h)n−2x + · · ·+ (x + h)xn−2 + xn−1
]= nxn−1.
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Table 1: Some elementary functions and their derivatives
f(x) f’(x)
c 0
x 1
x2 2x
1
x− 1
x2(x 6= 0)
√x
1
2√x
(x > 0)
x r rx r−1 (r 6= 0 and x r−1 is real)
|x | sgn(x) =x
|x |
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Leibniz Notation
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By the definition of derivative, we define the Leibniznotation as
dy
dx= lim
∆x→0
∆y
∆x.
The symbol ddx is a differential operator and is read as “the
derivative with respect to x of ...”
Remarks:
(i) d is not a variable, so do not cancel it out from the numeratorand denominator!
(ii) There are situations in which it may be convenient to considerdx and dy as separate quantities. However, they are NOT realnumbers and do not always follow the usual rules ofarithmetic.
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The following notations are equivalent:
dy
dx= y ′ = Dxy =
d
dxf (x) = f ′(x) = Dx f (x) = Df (x).
The following notations are equivalent:
dy
dx
∣∣∣x=x0
= y ′∣∣∣x=x0
= Dxy∣∣∣x=x0
=d
dxf (x)
∣∣∣x=x0
= f ′(x0) = Dx f (x0).
The symbol∣∣x=x0
is called an evaluation symbol, i.e., theexpression is evaluated at x = x0. For instance,
d
dxx3∣∣∣x=−4
= 3x2∣∣∣x=−4
= 3(−4)2 = 48.
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§2.3 Differentiation Rules
Theorem
If f is differentiable at x , then f is continuous at x .
Proof: Since f is differentiable at x , by definition we have
limh→0
f (x + h)− f (x)
h= f ′(x),
where f ′(x) is finite. Using the limit rules,
limh→0
(f (x + h)− f (x)) = limh→0
(f (x + h)− f (x)
h
)(h)
= f ′(x) · limh→0
h = 0.
This leads to limh→0
f (x + h) = f (x). That is, f is continuous at x .
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Linear Rules of Differentiation
Theorem
If f and g are differentiable at x , and if c is a constant, then thefunctions f + g , f − g and cf are all differentiable at x and
(f (x) + g(x))′ = f ′(x) + g ′(x),
(f (x)− g(x))′ = f ′(x)− g ′(x),
(cf (x))′ = cf ′(x).
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Proof: (i) For the sum rule, by definition we have
(f (x) + g(x))′ = limh→0
(f (x + h) + g(x + h))− (f (x) + g(x))
h
= limh→0
(f (x + h)− f (x)
h+
g(x + h)− g(x)
h
)= lim
h→0
f (x + h)− f (x)
h+ lim
h→0
g(x + h)− g(x)
h
= f ′(x) + g ′(x).
(ii) The proofs of the difference rule and constant rule are similarand are thus omitted.
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Product Rule of Differentiation
Theorem
[Product Rule] If f and g are differentiable at x , then theirproduct fg is also differentiable at x , and
(f (x)g(x))′ = f ′(x)g(x) + f (x)g ′(x).
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Proof: For the Product rule, by definition we have
(f (x)g(x))′ = limh→0
f (x + h)g(x + h)− f (x)g(x)
h
= limh→0
f (x + h)g(x + h)− f (x)g(x + h) + f (x)g(x + h)− f (x)g(x)
h
= limh→0
(f (x + h)− f (x)
hg(x + h) + f (x)
g(x + h)− g(x)
h
)= lim
h→0
f (x + h)− f (x)
hlimh→0
g(x + h) + f (x) · limh→0
g(x + h)− g(x)
h
= f ′(x)g(x) + f (x)g ′(x).
This proves the theorem.
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Example 6
Find the derivative of (x2 + 1)(x3 + 4) using and without using theProduct rule.
Solution: (i) Using the Product rule, we have
d
dx(x2 + 1)(x3 + 4) = 2x(x3 + 4) + (x2 + 1)(3x2)
= 5x4 + 3x2 + 8x .
(ii) Without using the Product rule, we have
d
dx(x2 + 1)(x3 + 4) =
d
dx(x5 + x3 + 4x2 + 4)
= 5x4 + 3x2 + 8x .
We note that the two solutions are the same.
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Quotient Rule of Differentiation
Theorem
[Quotient Rule] If f and g are differentiable at x , and ifg(x) 6= 0, then the quotient f /g is differentiable at x and(
f (x)
g(x)
)′=
f ′(x)g(x)− f (x)g ′(x)
g2(x).
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Proof: For the Quotient rule, by definition we have(f (x)
g(x)
)′
= limh→0
f (x + h)/g(x + h)− f (x)/g(x)
h
= limh→0
f (x + h)g(x)− f (x)g(x + h)
hg(x)g(x + h)
=1
g 2(x)limh→0
f (x + h)g(x)− f (x)g(x + h)
h
=1
g 2(x)limh→0
[f (x + h)− f (x)
hg(x)− f (x)
g(x + h)− g(x)
h
]=
f ′(x)g(x)− f (x)g ′(x)
g 2(x).
This proves the theorem.
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Reciprocal Rule of Differentiation
Theorem
[Reciprocal Rule] If g is differentiable at x and if g(x) 6= 0, thenthe reciprocal 1/g is also differentiable at x and(
1
g(x)
)′=−g ′(x)
g2(x).
Remarks: (i) The Reciprocal rule is a special case of the Quotientrule with f (x) = 1. (ii) By the Reciprocal rule, we can generalizethe Power rule to allow negative powers:
d
dxx−n = −nx−n−1.
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Example 7
Find the derivatives of (a) y =x2 + x + 1
x3; and (b) y =
x2
x + 1.
Solution:
(a) By the generalized Power rule, we have
dy
dx=
d
dx
(x−1 + x−2 + x−3
)= −(x−2 + 2x−3 + 3x−4).
(b) By the Quotient rule, we have
dy
dx=
(x2)′(x + 1)− x2(x + 1)′
(x + 1)2
=2x(x + 1)− x2
(x + 1)2
=x2 + 2x
(x + 1)2.
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Example 7
Find the derivatives of (a) y =x2 + x + 1
x3; and (b) y =
x2
x + 1.
Solution:
(a) By the generalized Power rule, we have
dy
dx=
d
dx
(x−1 + x−2 + x−3
)= −(x−2 + 2x−3 + 3x−4).
(b) By the Quotient rule, we have
dy
dx=
(x2)′(x + 1)− x2(x + 1)′
(x + 1)2
=2x(x + 1)− x2
(x + 1)2
=x2 + 2x
(x + 1)2.
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Example 7
Find the derivatives of (a) y =x2 + x + 1
x3; and (b) y =
x2
x + 1.
Solution:
(a) By the generalized Power rule, we have
dy
dx=
d
dx
(x−1 + x−2 + x−3
)= −(x−2 + 2x−3 + 3x−4).
(b) By the Quotient rule, we have
dy
dx=
(x2)′(x + 1)− x2(x + 1)′
(x + 1)2
=2x(x + 1)− x2
(x + 1)2
=x2 + 2x
(x + 1)2.
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§2.4 The Chain Rule
Theorem
[Chain Rule] If f (u) is differentiable at u = g(x), and g(x) isdifferentiable at x , then the composite function f ◦ g(x) = f (g(x))is differentiable at x , and
(f (g(x)))′ = f ′(g(x))g ′(x).
In Leibniz’s notation, by letting y = f (u) where u = g(x), we havey = f (g(x)) and
dy
dx=
dy
du
du
dx.
Remark: The proof of the Chain rule is not required for the course.
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The Chain rule is designed for composite functions and is themost commonly used rule of all the differentiation rules. Tounderstand the Chain rule, we can treat the Leibniz notationsdy/dx , dy/du, and du/dx as if they were quotients of twoquantities. The Chain rule can then be written as
dy
dx=
dy
du
du
dx.
Assume that y = f (u), u = g(v) and v = h(x). Theny = f (g(h(x))) and the Chain rule is
dy
dx=
dy
du
du
dv
dv
dx.
The Chain rule can be extended to composite functions withany finite number of intermediate functions.
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Example 8
Verify the Chain rule using the function y = 1/(x2 − 4).
Proof:
(i) By the Reciprocal rule, we have
dy
dx=−(x2 − 4)′
(x2 − 4)2=
−2x
(x2 − 4)2.
(ii) To verify the Chain rule, we let y = f (g(x)) with f (u) = 1/uand u = g(x) = x2 − 4. Then by the Chain rule,
dy
dx=
dy
du
du
dx=−1
u2(2x) =
−2x
(x2 − 4)2.
Thus, we obtain the same answer with the Chain rule.
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Example 8
Verify the Chain rule using the function y = 1/(x2 − 4).
Proof:
(i) By the Reciprocal rule, we have
dy
dx=−(x2 − 4)′
(x2 − 4)2=
−2x
(x2 − 4)2.
(ii) To verify the Chain rule, we let y = f (g(x)) with f (u) = 1/uand u = g(x) = x2 − 4. Then by the Chain rule,
dy
dx=
dy
du
du
dx=−1
u2(2x) =
−2x
(x2 − 4)2.
Thus, we obtain the same answer with the Chain rule.
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Example 8
Verify the Chain rule using the function y = 1/(x2 − 4).
Proof:
(i) By the Reciprocal rule, we have
dy
dx=−(x2 − 4)′
(x2 − 4)2=
−2x
(x2 − 4)2.
(ii) To verify the Chain rule, we let y = f (g(x)) with f (u) = 1/uand u = g(x) = x2 − 4. Then by the Chain rule,
dy
dx=
dy
du
du
dx=−1
u2(2x) =
−2x
(x2 − 4)2.
Thus, we obtain the same answer with the Chain rule.
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Example 9
Find the derivative of the function f (x) = (x +1
x)4.
Solution: Writing y = u4 where u = x + 1/x , we get
f ′(x) =dy
dx=
dy
du
du
dx
= (4u3)(1− 1
x2)
= 4(x +1
x)3(1− 1
x2).
Remark: those who are used to the chain rule will often simplywrite
f ′(x) = 4(x +1
x)3
[x +
1
x
]′= 4(x +
1
x)3(1− 1
x2).
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Example 10
Find the derivative of the function f (x) = (1 +√
2x3 + 1)3.
Solution: Let y = f (x). Then we have
y = u3, u = 1 +√v , v = 2x3 + 1,
dy
du= 3u2,
du
dv=
1
2√v,
dv
dx= 6x2.
By the Chain rule, we have
f ′(x) =dy
du
du
dv
dv
dx= 3u2 · 1
2√v· 6x2
=9x2(1 +
√2x3 + 1)3
√2x3 + 1
.
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Some Built-in Chain Rules
If u is a differentiable function of x , then we have the built-inChain rules for the following functions:
d
dx
(1
u
)=−1
u2
du
dx(Reciprocal Rule)
d
dx
√u =
1
2√u
du
dx(Square Root Rule)
d
dxur = rur−1 du
dx(General Power Rule)
d
dx|u| = sgn(u)
du
dx(Absolute Value Rule)
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Derivatives of Basic Elementary Functions
Recall that the elementary functions are built from the basicelementary functions and constants through composition(f ◦ g) and combinations using operations (+, −, × and /).
By the above rules of differentiation, we conclude that thederivatives of elementary functions are all available, given thatthe derivatives of basic elementary functions are provided.
We have learned the differentiation rules for power functions.In §2.5, we study the differentiation rules for trigonometricfunctions. The differentiation rules for (i) exponentialfunctions, (ii) logarithmic functions, and (iii) inversetrigonometric functions will be introduced in Chapter 3.
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§2.5 Derivatives of Trigonometric Functions
The trigonometric functions, especially sine and cosine, play avery important role in the mathematical modeling ofreal-world phenomena. In particular, they arise wheneverquantities fluctuate in a periodic way.
This section provides details in calculating the derivatives ofsine and cosine functions. The other four will then follow fromknown identities and the differentiation rules in Section 2.3.
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Some Special Limits
Theorem
The functions sin(θ) and cos(θ) are continuous at every value of θ.In particular, at θ = 0 we have
limθ→0
sin(θ) = sin(0) = 0
andlimθ→0
cos(θ) = cos(0) = 1.
Theorem
For the sine function, we have limθ→0
sin(θ)
θ= 1.
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Example 11
Show that limh→0
cos(h)− 1
h= 0.
Proof: Let θ = h/2, then h = 2θ. By §P.7, we have
cos(2θ) = cos2(θ)− sin2(θ) = 1− 2 sin2(θ).
This leads to
limh→0
cos(h)− 1
h= lim
θ→0
cos(2θ)− 1
2θ
= − limθ→0
sin2(θ)
θ
= − limθ→0
sin(θ)
θ· limθ→0
sin(θ)
= 0.
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Theorem
The derivative of the sine function isd
dxsin(x) = cos(x).
Proof: We have
d
dxsin(x) = lim
h→0
sin(x + h)− sin(x)
h
= limh→0
2 cos(x +h
2) sin(
h
2)
h
= limh→0
cos(x +h
2) · lim
h→0
sin(h/2)
h/2
= cos(x).
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Theorem
The derivative of the cosine function isd
dxcos(x) = − sin(x).
Proof: We have
d
dxcos(x) = lim
h→0
cos(x + h)− cos(x)
h
= limh→0
−2 sin(x +h
2) sin(
h
2)
h
= − limh→0
sin(x +h
2) · lim
h→0
sin(h/2)
h/2
= − sin(x).
Another simple proof is to apply for the Chain rule. Specifically,
d
dxcos(x) =
d
dxsin(
π
2− x) = − cos(
π
2− x) = − sin(x).
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Derivatives of Other Trigonometric Functions
tan(x) =sin(x)
cos(x)⇒ d
dxtan(x) = sec2(x)
cot(x) =cos(x)
sin(x)⇒ d
dxcot(x) = − csc2(x)
sec(x) =1
cos(x)⇒ d
dxsec(x) = sec(x) tan(x)
csc(x) =1
sin(x)⇒ d
dxcsc(x) = − csc(x) cot(x)
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§2.6 High-order Derivatives
Definition
The second derivative of a function is the derivative of itsderivative. If y = f (x), the second derivative is defined by
y ′′ = f ′′(x).
The following notations are referred to as the same:
y ′′ = f ′′(x) =d2y
dx2=
d
dx
d
dxf (x) =
d2
dx2f (x) = D2
x y = D2x f (x).
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Example 12
Find the second derivative of f (x) = 5x4 − 3x2 + 2x − 7.
Solution: The first derivative is
f ′(x) = 20x3 − 6x + 2.
The second derivative is then
f ′′(x) = 60x2 − 6.
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Definition
For any positive integer n, the nth derivative of a function isobtained from the function by differentiating successively n times.If the original function is y = f (x), the nth derivative is denoted by
dny
dxnor f (n)(x).
Specifically, we have
f (k)(x) =d
dx
[f (k−1)(x)
], k = 1, 2, · · · , n,
if f (0)(x), f (1)(x), · · · , f (n)(x) are all differentiable.
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Example 13
Calculate the first 3 derivatives of f (x) =√x2 + 1.
Solution: Note that f (x) = (x2 + 1)1/2. We have
f ′(x) =1
2(x2 + 1)−1/2(2x) = x(x2 + 1)−1/2,
f ′′(x) = (x2 + 1)−1/2 + x(−1
2)(x2 + 1)−3/2(2x)
= (x2 + 1)−3/2,
f (3)(x) = (−3
2)(x2 + 1)−5/2(2x) = −3x(x2 + 1)−5/2.
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By induction, we have the following higher order derivatives forsome important functions:
1) (xn)(k) =
n!
(n − k)!xn−k k = 1, 2, · · · , n
0 k > n
(2.5.1)
2) sin(n)(x) =
(−1)k−1 cos(x) n = 2k − 1
(−1)k sin(x) n = 2k(2.5.2)
3) cos(n)(x) =
(−1)k sin(x) n = 2k − 1
(−1)k cos(x) n = 2k(2.5.3)
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§2.8 The Mean-Value Theorem
Theorem (Mean-Value Theorem)
Suppose that the function f is continuous on the closed, finiteinterval [a, b] and that it is differentiable on the open interval(a, b). Then there exists a point c in the open interval (a, b) suchthat
f (b)− f (a)
b − a= f ′(c).
This says that the slope of the chord line joining the points(a, f (a)) and (b, f (b)) is equal to the slope of the tangent line tothe curve y = f (x) at the point (c , f (c)), so the two lines areparallel.
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The assumptions of the Mean-Value theorem are all necessaryfor the conclusion; see counterexamples on the next slide.
The Mean-Value theorem gives no indication of how manypoints C there may be on the curve between A and B wherethe tangent is parallel to AB.
The Mean-Value theorem gives us no information on how tofind the point c , even though such a point must exist. Thistype of theoretical tool is known as an existence theorem.
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Example 14
Show that sin(x) < x for all x > 0.
Proof:
(i) If x > π/2, then sin(x) ≤ 1 < π/2 < x .
(ii) If 0 < x ≤ π/2, then by the Mean-Value theorem, there existsa point c ∈ (0, π/2) such that
sin(x)
x=
sin(x)− sin(0)
x − 0=
d
dxsin(x)
∣∣∣x=c
= cos(c) < 1.
By (i) and (ii), sin(x) < x for all x > 0.
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Increasing and Decreasing Functions
Definition
Let f be a function defined on an interval I and x1, x2 be twopoints in I .
(a) If f (x1) < f (x2) whenever x1 < x2, then f is increasing on I .
(b) If f (x1) ≤ f (x2) whenever x1 < x2, then f is nondecreasingon I .
(c) If f (x1) > f (x2) whenever x1 < x2, then f is decreasing on I .
(d) If f (x1) ≥ f (x2) whenever x1 < x2, then f is nonincreasingon I .
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Increasing and Decreasing Functions
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Increasing and Decreasing Functions
Theorem
Let J be an open interval, and let I be an interval consisting of allpoints in J and possibly one or both of the endpoints of J.Suppose that f is continuous on I and differentiable on J.
(a) If f ′(x) > 0 for all x in J, then f is increasing on I .
(b) If f ′(x) < 0 for all x in J, then f is decreasing on I .
(c) If f ′(x) ≥ 0 for all x in J, then f is nondecreasing on I .
(d) If f ′(x) ≤ 0 for all x in J, then f is nonincreasing on I .
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Example 15
On what intervals is the function f (x) = x3 − 12x + 1 increasing?On what intervals is it decreasing?
Solution: The derivative of f is
f ′(x) = 3x2 − 12 = 3(x − 2)(x + 2).
Obviously, f ′(x) > 0 if x < −2 or x > 2; and f ′(x) < 0 if−2 < x < 2. Therefore, f is increasing on the intervals (−∞,−2)and (2,∞) and is decreasing on the interval (−2, 2).
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Example 16
On what intervals is the function f (x) =x
x2 + 1increasing? On
what intervals is it decreasing?
Solution: The derivative of f is
f ′(x) =(x)′(x2 + 1)− x(x2 + 1)′
(x2 + 1)2=
1− x2
(x2 + 1)2.
Obviously, f ′(x) > 0 if −1 < x < 1, and f ′(x) < 0 if x < −1 orx > 1. Therefore, f is increasing on the interval (−1, 1), and isdecreasing on the intervals (−∞,−1) and (1,∞).
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§2.9 Implicit Differentiation
If the dependent variable y can be written as an explicitformula of x , say y = f (x), then the derivative of y can beobtained using the differentiation rules.
Sometimes y cannot be written as a function of x explicitly.In this case, we say y is an implicit function of x . Write it asthe equation F (x , y) = 0.
The derivative dy/dx of implicit functions can be obtained bya technique called implicit differentiation. The idea is todifferentiate the equation with respect to x , regarding y as afunction of x having derivative dy/dx or y ′.
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Example 17: Calculate dy/dx at x = 3/5, where y is a functionof x defined implicitly by
x2 + y2 = 1.
Solution: The equation represents the unit circle, which is not thegraph of a function. However, it defines two differentiablefunctions y = f (x) and y = g(x), corresponding to the upper andlower branches of the graph:
f (x) =√
1− x2, g(x) = −√
1− x2.
When calculating dy/dx , one must specify which branch of thegraph we are interested in. This is usually done by also specifyingthe y value.
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−1.5 −1 −0.5 0 0.5 1 1.5
−1
−0.5
0
0.5
1(3/5, 4/5)
(3/5, −4/5)
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Example 17: Calculate dy/dx at the point (3/5, 4/5), where y isdefined implicitly by
x2 + y2 = 1.
Solution:
1 Check that the given point indeed lies on the curve:(35
)2+(
45
)2= 1.
2 Define the function
F (x) = x2 + (y(x))2 − 1,
which is identically zero because y(x) is assumed to satisfythe equation for all x near 3/5. We now differentiate F (x)using the chain rule:
dF
dx= 2x + 2y · dy
dx= 0.
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3 Isolate dy/dx to get
dy
dx= −x
y=⇒ dy
dx
∣∣∣∣x=3/5
= −3545
= −3
4.
Remarks:
We could have obtained the same answer by evaluatingf ′(3/5), where
f (x) =√
1− x2.
If we had used the point (3/5,−4/5) instead, the implicitfunction would have represented g(x) in the lower branch,where
dy
dx
∣∣∣∣x=3/5
= −35
−45
= +3
4.
Thus, the dependence on y in the implicit derivative reflectsthe fact that the equation defines more than one function.
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Example 18: Finddy
dxif y sin(x) = x3 + cos(y).
Solution: For this equation, y cannot be expressed as an explicitfunction of x . In what follows we use implicit differentiation.Specifically, we take the derivatives on both sides w.r.t x ,
d
dx(y sin(x)) =
d
dx(x3) +
d
dxcos(y).
This leads to
sin(x)dy
dx+ y cos(x) = 3x2 − sin(y)
dy
dx.
Solving this equation, we have
dy
dx=
3x2 − y cos(x)
sin(x) + sin(y).
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Example 19: Finddy
dxif xy2 = y − x .
Solution: Treat this as an implicit function. Taking the derivativeson both sides w.r.t x , we have
y2 + 2xydy
dx=
dy
dx− 1
This leads to
(1− 2xy)dy
dx= y2 + 1.
Further,
dy
dx=
y2 + 1
1− 2xy.
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§2.10 Antiderivatives
In the previous sections we have concerned with the problemof finding the derivative f ′ of a given function f . The reverseproblem, “given the derivative f ′, find f ”, is also interestingand important.
The reverse problem is related to integral calculus. We takea preliminary look at this problem in this section and willreturn to it with more detail in Chapter 5.
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Definition
A function F (x) is said to be an antiderivative of function f (x) if
F ′(x) = f (x)
for every x in the domain of f (x).
The process of finding antiderivatives is called anti-differentiationor indefinite integration.
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Example 20
Verify that F (x) = x3 + 5x is an antiderivative of f (x) = 3x2 + 5.
Solution: F (x) is an antiderivative of f (x) if
F ′(x) = f (x).
Differentiating F (x) we get
F ′(x) = 3x2 + 5 = f (x).
Remark: Let G (x) = x3 + 5x + 2. It is easy to verify that G (x) isalso an antiderivative of f (x). This shows that the antiderivativeof f (x) is not unique.
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Theorem (Fundamental Property of Antiderivatives)
If F (x) is an antiderivative of f (x), then the general expression forantiderivatives of f (x) is
F (x) + C ,
where C is an arbitrary constant.
Proof: Suppose that F (x) and G (x) are two antiderivatives of f (x). Letg(x) = G (x)− F (x). We have
g ′(x) = G ′(x)− F ′(x) = f (x)− f (x) = 0.
This leads to g(x) = C . Further,
G (x) = F (x) + g(x) = F (x) + C .
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Some examples
Function Antiderivatives
k kx + C
xa1
a + 1xa+1 + C (a 6= −1)
sin(x) − cos(x) + C
cos(x) sin(x) + C
1
cos2(x)tan(x) + C
1
sin2(x)− cot(x) + C
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Definition
The set of all antiderivatives of f (x) is called the indefiniteintegral, or simply the integral of f (x) with respect to x , denotedby ∫
f (x)dx = F (x) + C ,
where the symbol∫
is the integral sign, f (x) is the integrand ofintegral, x is the variable of integration, and C is the constantof integration.
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Example 21
Find
∫(3− x2)dx .
Solution: We have∫(3− x2)dx =
∫3 dx −
∫x2dx
= 3x − 1
3x3 + C .
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Example 22
Find
∫sin(2x)dx .
Solution: We seek a function whose derivative is sin(2x). Fromthe antiderivative tables, we guess a function of the typeg(x) = − cos(2x). But then the chain rule implies
g ′(x) = −(− sin(2x)) · 2 = 2 sin(2x).
Dividing both sides by 2 gives
d
dx
(1
2g(x)
)= sin(2x).
So the required antiderivative is∫sin(2x) dx = −1
2cos(2x) + C .
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