130 the derivative 2.2 Derivatives: Properties and Formulas This section begins with a look at which functions have derivatives. Then we’ll examine how to calculate derivatives of elementary combi- nations of basic functions. By knowing the derivatives of some basic functions and just a few differentiation patterns, you will be able to calculate the derivatives of a tremendous variety of functions. This section contains most—but not quite all—of the general differentiation patterns you will ever need. Which Functions Have Derivatives? A function must be continuous in order to be differentiable. It is vital to understand what this theo- Theorem: rem tells us and what it does not tell us: If a function is differentiable at a point, If a function is differentiable at a point then the function is automatically contin- then it is continuous at that point. uous there. If the function is continuous at a point, then the function may or may not be differentiable there. Proof. Assume that the hypothesis ( f is differentiable at the point c) is f (c + h) − f (c) true. Then lim must exist and be equal to f 0(c). We h!0 h want to show that f must necessarily be continuous at c, so we need to show that lim f (c + h) = f (c). h!0 It’s not yet obvious why we want to do so, but we can write: f (c + h) − f (c) f (c + h) = f (c) + · h h and then compute the limit of both sides of this expression: f (c + h) − f (c) lim f (c + h) = lim f (c) + · h h!0 h!0 h f (c + h) − f (c) = lim f (c) + lim · h h!0 h!0 h f (c + h) − f (c) = lim f (c) + lim · lim h h!0 h!0 h h!0 = f (c) + f 0(c) · 0 = f (c) Therefore f is continuous at c. We often use the contrapositive form of this theorem, which tells us about some functions that do not have derivatives. Contrapositive Form of the Theorem: If f is not continuous at a point then f is not differentiable at that point. 2.2 derivatives: properties and formulas 131 Example 1. Show that f (x) = bxc is not continuous and not differen- tiable at 2 (see margin figure). Solution. The one-sided limits lim bxc = 2 and lim bxc = 1 have x!2+ x!2− different values, so lim bxc does not exist. Therefore f (x) = bxc is not x!2 continuous at 2, and as a result it is not differentiable at 2. J Lack of continuity implies lack of differentiability, but the next exam- ples show that continuity is not enough to guarantee differentiability. Example 2. Show that f (x) = jxj is continuous but not differentiable at x = 0 (see margin figure). Solution. We know that lim jxj = 0 = j0j, so f is continuous at 0, but x!0 in Section 2.1 we saw that jxj was not differentiable at x = 0. J A function is not differentiable at a cusp or a “corner.” p 1 Example 3. Show that f (x) = 3 x = x 3 is continuous but not differen- tiable at x = 0 (see margin figure). p p p Solution. We can verify that lim 3 x = lim 3 x = 0, so lim 3 x = x!0+ x!0− x!0 p 3 0 1 − 2 1 0 = 0 so f is continuous at 0. But f (x) = x 3 = p , which is 3 3 3 x2 undefined at x = 0, so f is not differentiable at 0. J A function is not differentiable where its tangent line is vertical. Practice 1. At which integer values of x is the graph of f in the margin figure continuous? Differentiable? Graphically, a function is continuous if and only if its graph is “connected” and does not have any holes or breaks. Graphically, a function is differentiable if and only if it is continuous and its graph is “smooth” with no corners or vertical tangent lines. Derivatives of Elementary Combinations of Functions We now begin to compute derivatives of more complicated functions built from combinations of simpler functions. Example 4. The derivative of f (x) = x is D f (x) = 1 and the derivative of g(x) = 5 is D g(x) = 0. What are the derivatives of the elementary f combinations: 3 · f , f + g, f − g, f · g and ? g 132 the derivative Solution. The first three derivatives follow “nice” patterns: D(3 · f (x)) = D(3x) = 3 = 3 · 1 = 3 · D( f (x)) D( f (x) + g(x)) = D(x + 5) = 1 = 1 + 0 = D( f (x)) + D(g(x)) D( f (x) − g(x)) = D(x − 5) = 1 = 1 − 0 D( f (x)) − D(g(x)) yet the other two derivatives fail to follow the same “nice” patterns: D( f (x) · g(x)) = D(5x) = 5 but D( f (x)) · D(g(x)) = 1 · 0 = 0, and f (x) x 1 D( f (x)) 1 D = D = but = is undefined. g(x) 5 5 D(g(x)) 0 J The two very simple functions in the previous example show that, f D( f ) in general, D( f · g) 6= D( f ) · D(g) and D 6= . g D(g) Practice 2. For f (x) = 6x + 8 and g(x) = 2, compute the derivatives of f 3 · f , f + g, f − g, f · g and . g Main Differentiation Theorem: If f and g are differentiable at x, then: (a) Constant Multiple Rule: D(k · f (x)) = k · D( f (x)) (b) Sum Rule: D( f (x) + g(x)) = D( f (x)) + D(g(x)) (c) Difference Rule: D( f (x) − g(x)) = D( f (x)) − D(g(x)) (d) Product Rule: D( f (x) · g(x)) = f (x) · D(g(x)) + g(x) · D( f (x)) (e) Quotient Rule: f (x) g(x) · D( f (x)) − f (x) · D(g(x)) D = g(x) [g(x)]2 This theorem says that the simple patterns in the previous example for constant multiples of functions and sums and differences of func- tions are true for all differentiable functions. It also includes the correct patterns for derivatives of products and quotients of differentiable functions. 2.2 derivatives: properties and formulas 133 The proofs of parts (a), (b) and (c) of this theorem are straightforward, but parts (d) and (e) require some clever algebraic manipulations. Let’s look at some examples before tackling the proof. Example 5. Recall that D(x2) = 2x and D(sin(x)) = cos(x). Find d 2 D(3 sin(x)) and dx (5x − 7 sin(x)). Solution. Computing D(3 sin(x)) requires part (a) of the theorem with k = 3 and f (x) = sin(x) so D(3 · sin(x)) = 3 · D(sin(x)) = 3 cos(x), d while (5x2 − 7 sin(x)) uses part (c) of the theorem with f (x) = 5x2 dx and g(x) = 7 sin(x) so: d d d (5x2 − 7 sin(x)) = (5x2) − (7 sin(x)) dx dx dx d d = 5 · (x2) − 7 (sin(x)) dx dx = 5(2x) − 7(cos(x)) which simplifies to 10x − 7 cos(x). J d Practice 3. Find D(x3 − 5 sin(x)) and (sin(x) − 4x3). dx Practice 4. The table below gives the values of functions f and g, as well as their derivatives, at various points. Fill in the missing values for D(3 · f (x)), D(2 · f (x) + g(x)) and D(3 · g(x) − f (x)). x f (x) f 0(x) g(x) g0(x) D(3 f (x)) D(2 f (x) + g(x)) D(3g(x) − f (x)) 0 3 −2 −4 3 1 2 −1 1 0 2 4 2 3 1 Practice 5. Use the Main Differentiation Theorem to complete the table. f (x) g(x) x f (x) f 0(x) g(x) g0(x) D( f (x) · g(x)) D D g(x) f (x) 0 3 −2 −4 3 1 2 −1 1 0 2 4 2 3 1 d x3 Example 6. Determine D(x2 · sin(x)) and . dx sin(x) Solution. (a) Use the Product Rule with f (x) = x2 and g(x) = sin(x): Many calculus students find it easier to 2 remember the product rule in words: D(x · sin(x)) = D( f (x) · g(x)) = f (x) · D(g(x)) + g(x) · D( f (x)) “the first function times the derivative of = x2 · D(sin(x)) + sin(x) · D(x2) the second plus the second function times the derivative of the first.” = x2 · cos(x) + sin(x) · 2x = x2 cos(x) + 2x sin(x) 134 the derivative (b) Use the Quotient Rule with f (x) = x3 and g(x) = sin(x): d x3 d f (x) = dx sin(x) dx g(x) g(x) · D( f (x)) − f (x) · D(g(x)) = [g(x)]2 sin(x) · D(x3) − x3 · D(sin(x)) = [sin(x)]2 The quotient rule in words: “the bottom sin(x) · 3x2 − x3 · cos(x) time the derivative of the top minus the = top times the derivative of the bottom, all sin2(x) over the bottom squared.” 3x2 sin(x) − x3 · cos(x) = sin2(x) which could also be rewritten in terms of csc(x) and cot(x). J d 3t − 2 cos(x) Practice 6. Determine D((x2 + 1)(7x − 3)), and D . dt 5t + 1 x Now that we’ve seen how to use the theorem, let’s prove it. Proof.