1210-004 CALCULUS I
Review sheet for Midterm 2
February 26, 2014
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Derivatives and first applications
• Chain rule: this is a rule to compute derivatives of compositions of functions
f (g(x))0 = f 0 (g(x))g 0 (x)
Here are some examples:
(i)
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(2x5 −7)3
0
= (2x5 − 7)−3
0
= −3(2x5 − 7)−4 · 10x4 =
−30x4
(2x5 −7)4
(ii) sin(x2 ) = 2x cos(x2 )
0
(iii) sin(sin(x2 ))0 = cos(sin(x2 )) · sin(x2 ) = 2x cos(sin(x2 )) sin(x2 )
• Implicit differentiation. Most of the functions that we have been studying are explicit
functions y = f (x): this means that given any value of x, plugging into f we automatically
obtain a value for y. For example, given the parabola y = x2 , we know that a point with
x-coordinate x = 2 lies on the parabola if y = 22 = 4.
Functions might be given in a slightly more sophisticated fashion though. Consider the
curve given by the equation y 3 + 7y − x3 = 0 in which one cannot solve for y in terms of
x, so we don’t have an expression y = f (x). Given any value for x (say x = 2), plugging
it into the equation yields an equation in the variable y (for x = 2, we get y 3 + 7y = 8). It
turns out that for any x-value that one chooses, the resulting equation in y has a unique
solution (e.g., for x = 2, one sees that y = 1 is the unique y-value satisfying the equation).
In cases like these, we see that even though we don’t have an expression y = f (x), we are
still able to determine a unique y-value for any given x-value. We say that the equation
y 3 + 7y − x3 = 0 determines y as an implicit function of x.
The implicit differentiation method is a procedure for finding derivatives of functions
defined implicitly by equations.
(i) We start with an equation involving x and y which we know determines y as a function
of x
y 3 + 7y = x3
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(ii) We may thus think of y as a function of x y = f (x) (even though we don’t know an
expression for x). We replace y by f (x) in the equation and get
[f (x)]3 + 7f (x) = x3
(iii) Differentiate this equation with respect to x (using the chain rule)
3[f (x)]2 f 0 (x) + 7f 0 (x) = x2
(iv) Remember that y = f (x) and y 0 = f 0 (x) and bring y back into the equation
3y 2 y 0 + 7y 0 = 3x2
(v) The resulting equation will always be linear in y 0 , so we can solve for y 0 and obtain
y0 =
3x2
3y 2 + 7
If we want to obtain y 00 :
(i) Go back to step (iii) above and differentiate again with respect to x
3 · 2f (x)f 0 (x)f 0 (x) + 3[f (x)]2 f 00 (x) + 7f 00 (x) = 2x
(ii) Remember that y = f (x) and y 0 = f 0 (x) and bring y back into the equation
6y[y 0 ]2 + 3y 2 y 00 + 7y 00 = 2x
(iii) The resulting equation will always be linear in y 00 , so we can solve for y 00 and obtain
y 00 =
2x − 6y[y 0 ]2
3y 2 + 7
• Related rates. these are exercises in which a time-dependent variable y = y(t) is given
and in which ones wants to know how fast is y varying at a given time t0 (so one needs to
determine y 0 (t)). See the examples that we have worked through.
• Differentials and approximations. We defined the derivative of a function f at a point
x as the limit
f (x + h) − f (x)
f 0 (x) = lim
h→0
h
This means that when h is very small we have
f 0 (x) ∼
f (x + h) − f (x)
h
namely
f (x + h) ∼ f (x) + f 0 (x)h
If we denote h by dx, the above expression reads
f (x + dx) ∼ f (x) + f 0 (x)dx
which means that if we perturb x a small amount dx, we can approximate f (x + dx) by
solely looking at f (x) and f 0 (x).
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(i) For the function f (x) =
√
x, the above expression reads
√
so we can approximate
√
√
9.2 =
2
15.7 =
√
1
x + √ dx
2 x
9.02 as follows
√
9 + 0.2 '
(ii) Similarly, we can approximate
√
x + dx ∼
√
√
√
1
9 + √ · 0.2 = 16.0333
2 9
15.7 as follows
16 − 0.3 '
√
1
16 + √ · (−0.3) = 15.9625
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Applications of the derivative
• Absolute extreme points. These are the maximum and minimum values that a function
attains in a given interval. A continuous function f defined over a closed interval [a, b]
always attains both its maximum and minimum value over the interval.
1. THEOREM. If a point c in [a, b] is an extreme point (either the maximum or the
minimum) of a function f , then c must be a critical point, namely one of the following:
(i) An endpoint of the interval of definition of f : (a or b).
(ii) A singular point: namely a point c such that f 0 (c) does not exist.
(iii) A stationary point: namely, a point c such that f 0 (c) = 0.
2. Thus, the procedure for finding the extreme points of a function f defined over an
interval [a, b] is the following:
(i) Make a list containing all the critical points (endpoints, singular points and
stationary points).
(ii) Evaluate the function at each of these points.
(iii) The largest of these values will the the maximum of f and the smallest will be
the minimum (there might be more than one point where a function attains the
maximum value).
• Monotonicity and concavity. One may use derivatives in order to determine the behavior of a function over a given interval:
1. Suppose that f is a function whose derivative exists for every point inside an interval
I.
(i) If f 0 (x) > 0 for every x in I, then f is increasing over I.
(ii) If f 0 (x) < 0 for every x in I, then f is decreasing over I.
2. Suppose that f is a function whose first and second derivatives exists for every point
inside an interval I.
(i) If f 00 (x) > 0 for every x in I, then f is concave up over I (namely f 0 is increasing
over I).
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(ii) If f 00 (x) < 0 for every x in I, then f is concave down over I (namely f 0 is
decreasing over I)
3. An inflection point of a function f is a point c such that f is concave up on one side
of c, and concave down on the other. If c is an inflection point, then either f 00 (c) = 0
or c is a singular point of f . Thus, in order to find the inflection points of a function,
one can do the following:
(i) Make a list containing all the singular points (i.e. points c such that f 0 (c) does
not exist) and all the points c such that f 00 (c) = 0.
(ii) Check the sign of f 00 (x) for x < c and x > c
(iii) If the sign of f 00 (x) changes at c, then c is an inflection point.
• Local extreme points. A point c is a local extreme point (i.e. a local maximum or a
local minimum) if there is an interval (a, b) containing c such that f (c) is the maximum
or minimum value of f over (a, b) (that is, if we just look at f in a small neighborhood of
c, then c is the maximum or the minimum value).
1. First derivative test. Suppose that f is a continuous function defined over an
interval (a, b) which contains a point c and assume that f 0 (x) exists for every x in
(a, b) except for possibly c. Then:
(i) If f 0 (x) < 0 for x < c and f 0 (x) > 0 for x > c (namely, f goes from being
decreasing to being increasing at c), then c is a local minimum of f .
(ii)
(iii) If f 0 (x) > 0 for x < c and f 0 (x) < 0 for x > c (namely, f goes from being
increasing to being decreasing at c), then c is a local maximum of f .
2. Second derivative test. Suppose that f is a function defined over an interval (a, b)
which contains a point c and suppose that both f 0 (x) and f 00 (x) exist at every point
x in (a, b). Assume that f 0 (c) = 0.
(i) If f 00 (c) > 0 (so f is concave up), then f has a local minimum at c.
(ii) If f 00 (c) < 0 (so f is concave down), then f has a local maximum at c.
(iii) If f 00 (c) = 0, then this criterion doesn’t apply.
• Graphing functions. Given a function f defined over a interval or over the whole real
line (−∞, +∞), you can graph it precisely by following these steps:
1. Compute the value of the function at the endpoints of the interval of definition or
compute the limits of the function at those endpoints.
2. Determine the domain of f and classify the discontinuities (removable, jump or vertical asymptote).
3. Compute f 0 (x) and study its sign: this tells you over which intervals the function is
increasing or decreasing, the points for which f 0 vanishes (stationary points) and the
points at which f 0 does not exist (singular points).
4. Plot the stationary points and the singular points.
5. Using the first derivative criterion, you can then establish whether or not the stationary points and the singular points are local maximums or minimums.
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6. Compute f 00 (x) and study its sign: this tells you over which intervals the function
is concave up or down and also the points for which f 00 vanishes. If the sign of f 00
changes at the latter points or at the singular points, then those are inflection points.
7. Plot the inflection points.
8. Use all the information you have gathered to graph the function f .
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