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