We are looking for dydx, the rate of change of y with respect to x, at a point.. In the immediate vicinity of 1,√ 3 the curve looks like the graph of a differentiable function, despite t
Trang 16 Suppose y =f (x)g(x), where f (x) and g(x) are positive for all x Use logarithmic differ-entiation to finddydx Verify that this is the same result you would get had you used the Quotient Rule
7 If you felt so inclined, you could come up with a “rule” for taking the derivative of functions of the form f (x)g(x)where f (x) is positive You might call it “the Tower Rule” since you have a tower of functions, or you might think of a more descriptive name In any case, what would this rule be?
When using the process of logarithmic differentiation, we differentiate an equation in which
yis not explicitly expressed as a function of x Logarithmic differentiation is a special case
of the broader concept of implicit differentiation, a concept with far-reaching implications
and applications The basic idea is that we can finddydx even when y is not given explicitly
as a function of x We differentiate both sides of the equation that relates x and y, applying the Chain Rule to differentiate terms involving y because y varies with x
Implicit differentiation is an important concept; we’ll begin with a very straightforward example to illustrate what is going on
EXAMPLE 17.4 Consider the circle of radius 2 centered at the origin.5It is given by x2+ y2= 4 Find the
slope of the line tangent to the circle at the following points
(a) (1,√
3) (b) (1, −√3)
y
x
1 2
2
(1, –√3) (1, √3)
Figure 17.1 SOLUTION Although y is not a function of x, it can be expressed as two different functions of x
y =4 − x2(the top semicircle) and y = −4 − x2(the bottom semicircle) Each of these functions gives y explicitly as a function of x One approach is to differentiate the former expression to get information about the point (1,√
3) and the latter to deal with
5 This circle is the set of all points a distance 2 from the origin If (x, y) is a point on this circle, then the distance formula tells us that(x − 0) 2 + (y − 0) 2 = 2 And conversely, if (x, y) satisfies the equationx 2 + y 2 = 2, then (x, y) is a point on the circle Therefore, x 2
+ y2= 4 is the equation of the circle.
Trang 2(1, −√3) Do this on your own and compare your answers with those below We’ll take a more efficient approach
We are looking for dydx, the rate of change of y with respect to x, at a point The equation
x2+ y2= 4 implicitly gives a relationship between x and y Becausedxdyis a local measure
of a rate of change, we are interested in a bug’s-eye view rather than a bird’s-eye view of the curve Let’s peer at the point (1,√
3) through a magnifying glass In the immediate vicinity
of (1,√ 3) the curve looks like the graph of a differentiable function, despite the fact that a bird’s-eye view tells us it is not.6Because a derivative is a local measure of a rate of change, this is all that is really important We differentiate both sides of the equation treating y as if
it were a function of x We differentiate each side with respect to x because we are looking
fordxdy, the derivative of y with respect to x
d
dx[x2+ y2] =dxd [4]
Because we are treating y as if it were a function of x, when we differentiate y2we get 2ydydx Just as we needed to use the Chain Rule to find dxd [ln y] when we did logarithmic differentiation in Example 17.2, we need to use the Chain Rule here to evaluate dxdy2 Because y depends on x, what we are really trying to find here is dxd(mess)2 The Chain
Rule tells us that the derivative of (mess)2is 2(mess) · (mess) Therefore,
2ydy
dy
dx =−xy You may initially be startled that the formula for dydx involves not only x, but y also On
second thought, this should not be too surprising, because y is not a function of x A given
x-value may correspond to more than one y-value and therefore may have more than one slope associated with it We need to know both coordinates of a point on the curve, not just the x-value, to determine the slope of the tangent line For example, x = 1 at points (1,√3) and (1, −√3) on the circle; we need to specify a value for y to know which one is meant
It makes sense that our formula fordydx should involve both x and y
At (1,√
3), the slope of the tangent line is −1√
3, while at (1, −√3) the slope of the tangent line is√1
3 We’ve solved the original problem, but let’s look back at it one more time We found
dy
dx =−xy If we are looking at a point on the top semicircle where y =√4 − x2, we can write
dy
dx =√−x
4−x 2; if we were looking at a point on the bottom semicircle where y = −√4 − x2,
we can writedxdy= −x
−√
4−x 2
dy
dx is undefined when y = 0 This corresponds to the points (2, 0) and (−2, 0) on the circle Get out your magnifying glass again and take a good look at the curve in the immediate vicinity of each of these points No matter how much the curve is magnified,
6 The graph of the circle does not pass the vertical line test Any bird with enough height can see that.
Trang 3it does not look like the graph of a function Intuitively speaking, dydx is defined at a point
P, and we can find its value using implicit differentiation only if, under magnification, the curve around P looks like the graph of a differentiable function
EXAMPLE 17.5 The curve shown below is called the folium of Descartes It is the set of all points satisfying
the equation x3+ y3= 6xy The point (3, 3) lies on this curve; when we substitute x = 3 and y = 3 into the equation, both sides are equal Find the slope of a tangent line to
x3+ y3= 6xy, first in general, and then at the point (3, 3)
y
x
(3, 3)
x3 + y3 = 6xy
Folium of Descartes
Figure 17.2
SOLUTION We need to determinedydx Notice that we do not have an explicit formula for y in terms of
x In fact, a glance at the graph shows that y is not a function of x While it is possible to solve for y explicitly in terms of x using several functions, it is very difficult However, if
we magnify the curve right around the point (3, 3), it does look like the graph of a function,
so we can use implicit differentiation treating y as if it were a function of x and applying the Chain Rule where necessary
d
dx(x
3
+ y3) =dxd (6xy) Differentiate each side with respect to x,
applying the Chain Rule
3x2+ 3y2dydx = 6y + 6xdydx
In differentiating the 6xy on the right side of the equation, we have used the Chain Rule
in combination with the Product Rule Now we need to solve fordxdy The equation we have
is linear indxdy, so we use the standard strategy for solving linear equations
3x2+ 3y2dy
dx Bring all terms involvingdydx to one side 3y2dy
dx − 6xdxdy = 6y − 3x2 Factor out adydx dy
dx(3y2− 6x) = 6y − 3x2 Divide through to isolatedydx
dy
3y2− 6x dy
y2− 2x
Trang 4We now have a formula fordydx; this expression makes sense, provided y2 7
To find the slope of the line tangent to the folium of Descartes at the point (3, 3), substitute x = 3 and y = 3 into the expression fordydx, obtaining dydx= −1 This looks quite reasonable considering the symmetry of the graph around the line y = x
EXAMPLE 17.6 Find the slope of the tangent line to the curve 3y4+ 4xy2− 2x2= 9 at the point (2, 1)
SOLUTION We take the derivative of both sides with respect to x, thinking of y as if it were a function
of x
d
dx(3y4+ 4xy2− 2x2) =dxd (9) 12y3dy
dx + 4y2+ 8xydxdy − 4x = 0 Although we could solve fordxdy and then substitute in x = 2 and y = 1, because a general formula is not called for, it is algebraically much cleaner to substitute in the x- and y-values immediately after differentiation.8
12(1)3dy
dx + 4(1)2+ 8(2)(1)dydx − 4(2) = 0
28dy
dy
7 Thus, the slope at the point (2, 1) is 17
EXAMPLE 17.7 The equation 2x2+ 4xy + 3y2= 6 describes an ellipse9centered at the origin and rotated
as shown below Find the maximum and minimum values of y on this curve
y
x
√2
√3 –√2
–√3
Figure 17.3
SOLUTION We need to find the points at which the line tangent to the curve is horizontal, i.e., where
dy
dx = 0 To find dydx we use implicit differentiation
7 At the points on the folium of Descartes where y 2 = 2x the derivative is undefined y does not look like a differentiable function of x, no matter how much you magnify the area around the point in question For example, look at the point (0, 0).
8Don’t substitute in values of x and y before differentiating; you would just be taking the derivative of constants.
9 For information on conic sections, see Appendix E, Conic Sections.
Trang 5dx(2x2+ 4xy + 3y2) =dxd (6) 4x + 4y + 4xdxdy + 6ydydx = 0
dy
dy
2x + 3y
dy
dx= 0 where the numerator of the fraction is zero and the denominator is not simultaneously zero Simplifying −2x − 2y = 0 gives y = −x
We must check that the denominator is not zero at the same time Substituting y = −x
into the equation 2x + 3y = 0 gives x = 0 Thus, when x and y are both zero, dydx is not defined But the point (0, 0) is not on the ellipse, so we need not worry about it
y = −x is an entire line of points; we need to find out which of these points are also on the ellipse We find the points of intersection of the ellipse and the line by solving the two equations simultaneously
2x2+ 4xy + 3y2= 6
y = −x The simplest way to do this is to substitute the second equation into the first to eliminate the variable y and then solve for x
2x2+ 4xy + 3y2= 6
2x2− 4x2+ 3x2= 6
x2= 6
These are the two values of x at which the tangent line to the ellipse is horizontal Our original goal is to find the maximum and minimum y-values, so we can substitute these x-values back into the original equation Or, because we know that they will lie on the line
y = −x, and this is a much simpler equation, we can say that at x =√6, the minimum y-value of −√6 is attained and that at x = −√6, the maximum y-value of√
6 is attained
The Process of Implicit Differentiation
1.Decide which variable you want to differentiate with respect to (x if you wantdxdy, t if you wantdydt, etc.)
2.Differentiate both sides of the equation with respect to that variable Remember the Chain Rule!
Suppose you are differentiating with respect to t Distinguish between quantities that vary with t (treating them as functions of t) and those that are independent of t (treating them as constants) In particular, when looking fordy, think of y as a function of t
Trang 63.If necessary, solve to find a formula for the desired derivative If you only want to know the derivative at a specific point, substitute in the coordinates of that point before solving for the derivative you’re trying to find
A Brief Recap of Differentiating “with respect to” a Particular Variable
In practice, because a functional relationship between variables is not explicitly spelled out when we use implicit differentiation, it is necessary to think about what quantities change with which other variables Suppose, for instance, that we will differentiate both sides of an equation with respect to the variable time t; we must establish which quantities vary with respect to time and which do not We’ll clarify this by means of an example
EXAMPLE 17.8 Let P be the pressure under which a gas is kept, V be the volume of the gas, and T be
temperature measured on the absolute (or Kelvin) scale Then the combined gas law tells
us thatP VT = K, where K is a constant.10 (a) Suppose temperature is kept constant Express the rate of change of volume with respect
to pressure
(b) Suppose volume is kept constant Express the rate of change of temperature with respect
to pressure.11 (c) Suppose volume is kept constant but pressure and temperature change with time What
is the relationship between the change in pressure with respect to time and the change
in temperature with respect to time?
(d) Suppose pressure, temperature, and volume all change with time How are these rates
of change related?
SOLUTION (a) Here we are thinking of P as the independent variable and looking for dVdP V varies
with P , so we treat it as a function of P T is treated as a constant in this part of the problem K is also a constant We can emphasize this by writing
P
TV (P ) = K
We want to finddVdP One option is to solve for V in terms of P to get
To find out how V varies with P we differentiate with respect to P , obtaining
d
dP[V ] =dPd [KT P−1]
10 Here’s a little history for the chemists among you In 1660 Robert Boyle published his gas law stating that if temperature
is kept constant, then the product of the pressure and the volume is a constant Later Jacques Charles looked at the relationship between volume and temperature when pressure is kept constant The absolute (Kelvin) temperature scale allows Charles’s law
to be written V /T = C 1 , where T is the temperature on the absolute scale and the constant C 1 depends on the pressure and the mass of gas present In 1802 Joseph Guy-Lussac’s investigation of gases yielded the result that P /T = C 2 , where C 2 is a constant Combining Boyle’s, Charles’s, and Guy-Lussac’s laws gives the more general gas law, P V /T = C 3 , where the value of the constant
C 3 depends on the amount of gas present and T is the absolute (or Kelvin) temperature I found this law handy on Christmas day
1996 when baking a cake for a potluck dinner in the Ecuadoran Andes at an altitude of 2530 meters (about 1.5 miles above sea level).
11 This was the goal in the cake baking mentioned in the previous footnote The pressure had dropped due to the high altitude.
I was trying to figure out how the baking temperature should change in order to counterbalance this.
Trang 7dP = KT (−1)P−2=−KTP2 12 Another option is to differentiate implicitly
d dP
P
T V (P )
=dPd K, so 1
T
= 0 P
T
dV
dP =−V (P )T Solving fordV
dP gives dV
dP =−V (P )P This is equivalent to the previous answer, because
V (P ) =KTP
(b) Here we are thinking of P as the independent variable and looking for dTdP Now V
is treated as a constant and T as a function of P We can emphasize this by writing
P V
T (P )= K The simplest strategy here is to solve for T explicitly and then differentiate with respect to P
T (P ) =P V
KP, where V and K are both constant Therefore,dT
K (c) Here we are thinking of time t as the independent variable; T and P vary with time but
we consider both V and K as constants We differentiate implicitly
It may be helpful to write
P (t ) · V
to emphasize that P and T vary with time
We will differentiate the equation with respect to t Our job will be easier if we rewrite the equation as
V · P (t) = KT (t) because we won’t need the Product and the Quotient Rules
Differentiating with respect to t gives
dt = KdTdt
dt =KV dTdt
If we like, we can eliminate V entirely, because V = KTP
12Some remarks about notation:
d
dP is an operator; it is a symbol that represents the operation of differentiating with respect to P dV is an expression, in the same way that P + V is an expression.
dV = lim "P →0"V, where "V = V (P + "P ) − V (P ) The notation dV reminds us that we are differentiating V with respect to P Here it reminds us that we are looking at the change in volume induced by a change in pressure.
Trang 8dt = K · KTP ·dTdt
dt =PT dTdt (d) Again we think of time t as the independent variable Here P , V , and T vary with time
It may be helpful to write
V (t ) · P (t) = K · T (t)
We’ll differentiate with respect to t, so we’ll need to use the Product Rule on the left-hand side
V (t ) ·dP
dt · P (t) = K ·dT
dt
We can eliminate K completely, replacing it by P VT
V (t ) ·dPdt +dVdt · P (t) =P VT · dTdt Observe that if we know P ,V , and T at a certain instant, then knowing two of the rates of change at that moment allows us to determine the third
REMARKNotice that the notation V can be ambiguous Is itdVdPordVdt? We use it only when
it is clear from context what is meant The notation V can also be ambiguous The second derivative of V with respect to P , d
dP
dV dP
can be written as d2V
dP 2 The second derivative
of V with respect to t,dtd dVdtcan be written as d2V
dt 2
P R O B L E M S F O R S E C T I O N 1 7 3
1 The equation x2+ y2= 169 describes a circle with radius 13 centered at the origin (a) Solve explicitly for y in terms of x Is y a function of x?
(b) Differentiate your expression(s) from part (a) to finddydx (c) Now use implicit differentiation on the original equation to find dydx (d) Which method of differentiation (that used in part(a), or that used in part (b)) was easier? Why?
(e) What is the slope of the tangent line to the circle at the point (5, 12)? At (5, −12)?
2 Which is larger, the slope of the tangent line to x2+ y2= 25 at the point (4, −3) or the slope of the tangent line to x2+ 4y2= 25 at the point (4, −3/2)? You can answer this analytically (find the two slopes), or you can answer this by looking at the graphs
of the ellipse and the circle and sketching the tangent to each at the designated points (To sketch the ellipse, look at the x- and y-intercepts.)
3 Consider the equation x2y + xy2+ x = 1 Finddxdy at all points where x = 1
4 Find the slopes of the tangent lines to (x − 2)2+ (y − 3)2= 25 at the two points where
x = 6
Trang 95 Find the equation of the line tangent to the curve x3+ y3− 3x2y2+ 1 = 0 at the point (1, 1)
6 Find dydx for the curve x3+ 3y + y2= 6 What is the slope at (2, −1)? At what points
is the slope zero?
7 At what points (if any) is the tangent line to the curve 3x2+ 6xy + 8y2= 8 vertical?
8 (a) Graph the ellipse 4(x − 1)2+ 9(y − 3)2= 36 (Sketch 4x2+ 9y2= 36 by looking for x- and y-intercepts and then shift the graph horizontally and vertically as appropriate.)
(b) Find the slope of the line tangent to the ellipse at the point (1, 5) in two ways First solve for y explicitly (using the appropriate half of the ellipse) and find dydx Then
do the problem by differentiating implicitly
9 Consider the circle given by x2+ y2= 4
(a) Show thatdxdy= −xy
(b) Show thatd2y
dx 2 =dxd(−xy) = −y43 Explain your reasoning completely
10 The equation 2(x2+ y2)2= 25(x2− y2)gives a curve that is known as a lemniscate Find the slope of the tangent line to the lemniscate at the point (−3, 1)
x y
Lemniscate: 2(x2+ y2 ) 2 = 25(x2 – y2 )
11 Find an equation for the tangent line to y2= x3(3 − x) at the point (1, 2) What can you say about the tangent line to this curve at the point (3, 0)?
x y
y2 = x3 (3 – x)
12 Find dydx
(a) 3x2+ 6y2+ 3xy = 10 (b) (x − 2)3· (y − 2)3= 1 (c) xy2+ 2y = x2y + 1 (d) (x2y3+ y)2= 3x (e) exy
= y2 (f) x ln(xy3) = y2 (g) ln(xy) = xy2
Trang 1013 (a) Sketch a graph of (x − 2)2− (y − 3)2= 25 (Sketch x2− y2= 25 by looking for x- and y-intercepts and then shift your graph vertically and horizontally as appropriate.)
(b) Find the equation of the tangent line to the graph at each of the following points
14 Find an equation for the tangent line to x2/3+ y2/3= 5 at the point (8, 1)
x y
x + y = 5: astroid
53/
53/
2 / 2 /
15 Match the equations with the appropriate graphs (These are graphs of conic sections
To learn more about conic sections, refer to Appendix E.)
(a) x2
5 + y52 = 8 (b) 3x2− 3y2= 27 (c) 3x2+ y2= 12 (d) x2+ 3y2= 12 (e) −x2+ 3y2= 12
y
x A
y
x
C
y
x
B
y
x E
y
x
D
RELATED RATES OF CHANGE
In Sections 17.2 and 17.3 we introduced logarithmic and implicit differentiation The former
is particularly useful when we want to differentiate something of the form [f (x)]g(x)and the latter when it is either unappealing or impossible to solve for the variable we wish
to differentiate In fact, we can get a great deal more mileage from the two fundamental principles we used
Principle (i):If two expressions are equal, their derivatives are equal
Principle (ii):Whether y is given implicitly or explicitly in terms of x, we can differ-entiate an expression in y with respect to x by treating y the same way we would treat g(x)and using the Chain Rule