**Chapter 3**

October 16 2016

**Chapter 3 The Computation of Derivatives**
**3.1 Derivatives of Polynomials**
- differential calculus, the calculus of derivatives, takes its special flavour and importance from its many applications to the physical, biological, and social sciences
- it would be pleasant to plunge into these applications, but it is more efficient to learn how to calculate derivatives with speed and accuracy
- previously, differentiation was based directly on the limit definition of the derivative:
- f'(x) = lim dx>0 f(x+dx) - f(x) / dx, or,
- dy/dx = d/dx f(x) = lim dx>0 dy / dx
- the approach is rather slow and clumsy
- now we develop a small number of formal rules that will enable us to differentiate large classes of functions quickly, by purely mechanical procedures
- first we start with polynomials, later we can cope with messy algebraics
- recall that a polynomial in x is the sum of constant multiples of powers of x in which each exponent is zero or a positive integer
- P(x) = anxn + an-1xn-1 + ... + a1x + a0
- the way a polynomial is put together out of simpler pieces suggests the differentiation rules we now discuss
**1. The derivative of a constant is zero**
- d/dx c = 0
- the geometric meaning is that a horizontal straight line y = f(x) = c has zero slope
- to prove: expand with three step rule
**2. If n is a positive integer, then d/dx xn = nxn-1**
- the derivative of xn is obtained by bringing the exponent n down in front as a coefficient, then subtracting 1 from it to form the new exponent
- d/dx x2 = 2x, d/dx x3 = 3x2
- to prove: expand with three step rule and binomial theorem
**3. If c is a constant and u = f(x) is a differentiable function of x, then d/dx cu = c du/dx**
- d/dx cxn = cnxn-1
**4. If u = f(x) and v = g(x) then d/dx (u + v) = du/dx + du/dx**
- it is now easy to differentiate any polynomial
- even though the letters x and y are often used for the independent and dependent variables
- any letters are ok (e.g. ds/dt da/ds for speed and acceleration)
**3.2 The Product and Quotient Rules**
- since the derivative of a sum is the sum of the derivatives, it is natural to guess that the derivative of a product might equal the product of the derivatives
- this is not true, for example x3 x4 = x7 so the derivative of the product is 7x6 but the product of the individual derivatives is 3x2 4x3 = 12x5
- the correct formula for differentiating products is rather surprising
**5. The product rule - d/dx (uv) = u dv/dx + v dv/dx**
- the derivative of a product of two functions is the first times the derivative of the second plus the second times the derivative of the first
- to prove: y = ux expand by the three-step rule
- when both factors are polynomials, the product rule is not needed because the product of two polynomials is also a polynomial
- but more complex situations lie ahead - in which factors are often different types of functions - it will be clear that the product rule is indispensible
**6. The quotient rule - d/dx (u/v) = v du/dx - u dv/dx / v2**
- at all values of x where v is not 0
- the derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the denominator squared
- to prove: expand with the three-step rule
- the quotient rule enables us to extend rule 2 to cases where n is a negative integer
- d/dx x-1 = -1x-2
- d/dx x2 = -2x-3
**3.3 Composite Functions and the Chain Rule**
- lets consider the problem of differentiating this function
- y = (x3 + 2)5
- we can do this by expanding but it is bothersome, it is much better to develop the chain rule
- we decompose the function into single pieces by:
- y = u5 where u = x3 + 2
- we can reconstruct the functions out of these pieces by creating a composite function, a function of a function:
- y = f(u) where u = g(x) y = f(g(x))
- the chain rule is about decomposing a function into simpler functions and using the presumably simpler derivatives of these simpler functions
**7. The chain rule dy/dx = dy/du du/dx**
- its intuitive content is easy to grasp if we think of derivatives as rates of change
- if y changes a times as fast as u
- and u changes b times as fast as x
- then y changes ab times as fast as x
- if a car travels twice as fast as a bicycle and the bicycle is three times as fast as a walking man, then the car travels ten times as fast as the man
- so dy/dx = dy/du du/dx = 5u4 3x2
- the chain rule is indispensable for almost all differentiations above the simplest level
- an important special case is:
- d/dx ()n = n()n-1 d/dx ()
- where any differentiable function of x can be inserted in the parentheses
- if we denote the function by u, the formula can be written as follows
**8. The power rule d/dx un = nun-1 dy/dx**
- at this state we know the exponent n is allowed to be any integer
- later we will see it is also valid for all fractional exponents
- in Chapter 4 we will be using derivatives as tools in many concrete problems of science and geometry, and it will be clear that it is worth a little extra effort to put the derivatives we calculate into their simplest possible forms
**3.4 Implicit Functions and Fractional Exponents**
- most of the functions we have met so far have been of the form y = f(x), in which y is expressed directly, or explicitly, in terms of x
- in contrast to this, it often happens that y is defined as a function of x by means of an equation:
- F(x, y) = 0
- which is not solved for y but in which x and y are more or less entangled with each other
- when x is given a suitable numerical value, the resulting equation usually determines one or more corresponding values of y
- in such a case we say that the equation determines y as one or more implicit functions of x
- xy = 1 determines one implicit function of x, which can be written explicitly as y = 1/x
- x2 + y2 = 25 determines two implicit functions of x, which can be written explicitly as y = +/- sqrt (25 - x2)
- as we know, these two functions are the upper and lower halves of the circle of radius 5
- it is rather surprising that we can often calculate the derivative dy/dx of an implicit function without first solving the equation for y
- we start by differentiating the given equation through with respect to x, using the chain rule (or power rule) and consciously thinking of y as a function of x wherever it appears
- for example, y3 is treated as the cube of a function of x and its derivative d/dx y3 = 3y2 dy/dx
- and x3y4 is thought of as the product of two functions of x and its derivative is d/dx x3y4 = x3 4y3 dy/dx + y4 3x2
- to complete the process, we solve the resulting equation for dy/dx as the unknown, this method is called implicit differentiation
- for example, xy = 1, d/dx both sides: x dy/dx + y = 0, dy/dx = -y/x
- it is apparent that implicit differentiation usually gives an expression for dy/dx in terms of both x and y, instead of x alone
- however, in many cases, this is not a real disadvantage
- for instance, if we want the slope of the tangent to the graph of the equation at a point (x0, y0), all we have to do is substitute the values for x0 and y0 in
- we can use implicit differentiation to show the power rule works for all integers of n as well as for all fractional exponents n
- for example, d/dx x 1/2 = 1/2 x -1/2
**3.5 Derivatives of Higher Order**
- the derivative of y = x4 is clearly y' = 4x3
- but 4x3 can also be differentiated, yielding 12x2
- it is natural to denote this function y'' and call it the second derivative of the original function
- several notations are in common use for these higher-order derivatives
- f''(x) y'' d2y/dx2 d2/dx2 f(x)
- d2y/dx2 = d/dx (dy/dx)
- what are the uses of these higher derivatives?
- in geometry, the sign of f''(x) tells us whether the curve y = f(x) is concave up or concave down
- also, in a later chapter this qualitative interpretation of the second derivative will be refined into a quantitative formula for the curvature of the curve
- in physics, second derivatives are of very great importance
- if s = f(t) gives the position of a moving body at time t,
- then we know the first and second derivatives of this position function are the velocity and acceleration of the body at time t
- the central role of acceleration arises from Newton's second law of motion, which states that the acceleration of a moving body is proportional to the force acting on it
- the basic problem of Newtonian dynamics is to use calculus to deduce the nature of the motion from the given force
- derivatives of higher order than the second do not have such fundamental geometric or physical interpretations
- however, as we shall see later, these derivatives have their uses too, mainly in connection with expanding functions into infinite series