Exponential and Log, Logarithmic Differentiation, Hyperbolic Functions

Lecture 6 Sept. 19, 2006 18.01 Fall 2006 Lecture 6: Exponential and Log, Logarithmic Differentiation, Hyperbolic Functions Taking the derivatives of exponentials and logarithms Background We always assume the base, a, is greater than 1. a0 = 1; ax1 +x2 x (ax1 ) 2 p aq a1 = a; a2 = a · a; ... ax1 ax2 a x1 x2 √ q ap (where p and q are integers) = = = To define ar for real numbers r, fill in by continuity. Today’s main task: find We can write d x a dx d x ax+Δx − ax a = lim Δx→0 dx Δx We can factor out the ax : aΔx − 1 aΔx − 1 ax+Δx − ax = lim ax = ax lim Δx→0 Δx→0 Δx→0 Δx Δx Δx lim Let’s call aΔx − 1 Δx→0 Δx We don’t yet know what M (a) is, but we can say M (a) ≡ lim d x a = M (a)ax dx Here are two ways to describe M (a): 1. Analytically M (a) = d x dx a at x = 0. � a0+Δx − a0 d x �� Indeed, M (a) = lim = a Δx→0 Δx dx �x=0 1 Lecture 6 Sept. 19, 2006 18.01 Fall 2006 ax M(a) (slope of ax at x=0) Figure 1: Geometric definition of M (a) 2. Geometrically, M (a) is the slope of the graph y = ax at x = 0. The trick to figuring out what M (a) is is to beg the question and define e as the number such that M (e) = 1. Now can we be sure there is such a number e? First notice that as the base a increases, the graph ax gets steeper. Next, we will estimate the slope M (a) for a = 2 and a = 4 geometrically. Look at the graph of 2x in Fig. 2. The secant line from (0, 1) to (1, 2) of the graph y = 2x has slope 1. Therefore, the slope of y = 2x at x = 0 is less: M (2) < 1 (see Fig. 2). 1 1 Next, look at the graph of 4x in Fig. 3. The secant line from (− , ) to (1, 0) on the graph of 2 2 y = 4x has slope 1. Therefore, the slope of y = 4x at x = 0 is greater than M (4) > 1 (see Fig. 3). Somewhere in between 2 and 4 there is a base whose slope at x = 0 is 1. 2 Lecture 6 Sept. 19, 2006 18.01 Fall 2006 y=2x slope = 1 (1,2) e lin t n eca s slope M(2) Figure 2: Slope M (2) < 1 y=4x (4) eM p o l s (1,0) (-1/2, 1/2) Figure 3: Slope M (4) > 1 3 e t lin secan Lecture 6 Sept. 19, 2006 18.01 Fall 2006 Thus we can define e to be the unique number such that M (e) = 1 or, to put it another way, eh − 1 =1 h→ 0 h lim or, to put it still another way, d x (e ) = 1 dx What is d x (e )? dx at x = 0 We just defined M (e) = 1, and d x (e ) = M (e)ex . So dx d x (e ) = ex dx Natural log (inverse function of ex ) To understand M (a) better, we study the natural log function ln(x). This function is defined as follows: If y = ex , then ln(y) = x (or) If w = ln(x), then ex = w Note that ex is always positive, even if x is negative. Recall that ln(1) = 0; ln(x) < 0 for 0 < x < 1; ln(x) > 0 for x > 1. Recall also that ln(x1 x2 ) = ln x1 + ln x2 Let us use implicit differentiation to find d ln(x). dx ew d w (e ) dx d w dw (e ) dw dx w dw e dx dw dx = = w = ln(x). We want to find x d (x) dx = 1 = 1 = 1 1 = ew x d 1 (ln(x)) = dx x 4 dw . dx Lecture 6 Sept. 19, 2006 Finally, what about 18.01 Fall 2006 d x (a )? dx There are two methods we can use: Method 1: Write base e and use chain rule. Rewrite a as eln(a) . Then, � �x ax = eln(a) = ex ln(a) That looks like it might be tricky to differentiate. Let’s work up to it: d x e dx ex = and by the chain rule, d 3x e dx 3e3x = Remember, ln(a) is just a constant number– not a variable! Therefore, d (ln a)x e dx = or (ln a)e(ln a)x d x (a ) = ln(a) · ax dx Recall that d x (a ) = M (a) · ax dx So now we know the value of M (a): M (a) = ln(a). Even if we insist on starting with another base, like 10, the natural logarithm appears: d x 10 = (ln 10)10x dx The base e may seem strange at first. But, it comes up everywhere. After a while, you’ll learn to appreciate just how natural it is. Method 2: Logarithmic Differentiation. The idea is to find u = f (x). Since u = f and d d f (x) by finding ln(f (x)) instead. Sometimes this approach is easier. Let dx dx � � d d ln(u) du 1 du ln(u) = = dx du dx u dx du = f � , we can also write dx (ln f )� = f� f or 5 f � = f (ln f )� Lecture 6 Sept. 19, 2006 18.01 Fall 2006 Apply this to f (x) = ax . ln f (x) = x ln a =⇒ d d d ln(f ) = ln(ax ) = (x ln(a)) = ln(a). dx dx dx (Remember, ln(a) is a constant, not a variable.) Hence, d f� d x (ln f ) = ln(a) =⇒ = ln(a) =⇒ f � = ln(a)f =⇒ a = (ln a)ax dx f dx Example 1. d x (x ) = ? dx With variable (“moving”) exponents, you should use either base e or logarithmic differentiation. In this example, we will use the latter. f ln f = = � (ln f ) = (ln f )� = xx x ln x � � 1 1 · (ln x) + x = ln(x) + 1 x f� f Therefore, f � = f (ln f )� = xx (ln(x) + 1) If you wanted to solve this using the base e approach, you would say f = ex ln x and differentiate it using the chain rule. It gets you the same answer, but requires a little more writing. � Example 2. Use logs to evaluate lim k→∞ 1+ 1 k �k . Because the exponent k changes, it is better to find the limit of the logarithm. �� lim ln k→∞ We know that 1 1+ k �k � �� �k � � � 1 1 ln 1 + = k ln 1 + k k � � 1 → 0. This expression has two competing parts, which balance: k → ∞ while ln 1 + k �� � � �k � � � ln 1 + k1 1 1 ln(1 + h) 1 ln 1 + = k ln 1 + = = (with h = ) 1 k k h k k Next, because ln 1 = 0 �� �k � 1 ln(1 + h) − ln(1) ln 1 + = k h 6 Lecture 6 Take the limit: h = Sept. 19, 2006 18.01 Fall 2006 1 → 0 as k → ∞, so that k � ln(1 + h) − ln(1) d � = ln(x)� =1 h→0 h dx x=1 lim In all, � �k 1 lim ln 1 + = 1. k→∞ k � �k 1 We have just found that ak = ln[ 1 + ] → 1 as k → ∞. k � �k 1 If bk = 1 + , then bk = eak → e1 as k → ∞. In other words, we have evaluated the limit we k wanted: � �k 1 lim 1 + =e k→∞ k Remark 1. We never figured out what the exact numerical value of e was. Now we can use this limit formula; k = 10 gives a pretty good approximation to the actual value of e. Remark 2. Logs are used in all sciences and even in finance. Think about the stock market. If I say the market fell 50 points today, you’d need to know whether the market average before the drop was 300 points or 10, 000. In other words, you care about the percent change, or the ratio of the change to the starting value: f � (t) d = ln(f (t)) f (t) dt 7