Lecture 5 Notes: The Derivatives of Exponential and Logarithm Functions

18.01 Calculus Jason Starr Fall 2005 Lecture 5. September 16, 2005 Homework. Problem Set 2 Part I: (a)–(e); Part II: Problem 2. Practice Problems. Course Reader: 1I­1, 1I­4, 1I­5 1. Example of implicit differentiation. Let y = f (x) be the unique function satisfying the equation, 1 1 + = 2. x y What is slope of the tangent line to the graph of y = f (x) at the point (x, y) = (1, 1)? Implicitly differentiate each side of the equation to get, � � � � d 1 d 1 d(2) + = = 0. dx x dx y dx Of course (1/x)� = (x−1 )� = −x−2 . And by the rule d(un )/dx = nun−1 (du/dx), the derivative of 1/y is −y −2 (dy/dx). Thus, dy −x−2 − y −2 = 0. dx Plugging in x equals 1 and y equals 1 gives, −1 − 1y � (1) = 0, whose solution is, y � (1) = −1 . In fact, using that 1/y equals 2 − 1/x, this can be solved for every x, dy 1 1 1 = (x−2 )/(y −2 ) = 2 · = . 2 dx x (2 − 1/x) (2x − 1)2 2. Rules for exponentials and logarithms. Let a be a positive real number. The basic rules of exponentials are as follows. Rule 1. If ab equals B and ac equals C, then ab+c equals B · C, i.e., ab+c = ab · ac . Rule 2. If ab equals B and B d equals D, then abd equals D, i.e., (ab )d = abd . If ab equals B, the logarithm with base a of B is defined to be b. This is written loga (B) = b. The function B → loga (B) is defined for all positive real numbers B. Using this definition, the rules of exponentiation become rules of logarithms. 18.01 Calculus Jason Starr Fall 2005 Rule 1. If loga (B) equals b and loga (C) equals c, then loga (B · C) equals b + c, i.e., loga (B · C) = loga (B) + loga (C). Rule 2. If loga (B) equals b and B d equals D, then loga (D) equals d loga (B), i.e., loga (B d ) = d loga (B). Rule 3. Since logB (D) equals d, an equivalent formulation is loga (D) equals loga (B) logB (D), i.e., loga (D) = loga (B) logB (D). 3. The derivative of ax . Let a be a positive real number. What is the derivative of ax ? Denote the derivative of ax at x = 0 by L(a). It equals the value of the limit, ah − 1 . h→0 h L(a) = lim Then for every x0 , the derivative of ax at x0 equals, ax0 +h − ax0 . h→0 h lim By Rule 1, ax0 +h equals ax0 ah . Thus the limit factors as, ax 0 ah − ax 0 = ax0 lim ah − 1h. h→0 h→0 h lim Therefore, for every x, the derivative of ax is, d(ax ) = L(a)ax . dx What is L(a)? To figure this out, consider how L(a) changes as a changes. First of all, (ab )h − 1 . h→0 h L(ab ) = lim By Rule 2, (ab )h equals abh . So the limit is, abh − 1 abh − 1 = b lim . h→0 h→0 h bh L(ab ) = lim Now, inside the limit, make the substitution that k equals bh. As h approaches 0, also k approaches 0. So the limit is, ak − 1 = bL(a). L(ab ) = b lim k→0 k 18.01 Calculus Jason Starr Fall 2005 This is very similar to Rule 2 for logarithms. Choose a number a0 bigger than 1, say a0 = 2. Then for every positive real number a, a = ab0 where b = loga0 (a). Thus, L(a) = L(ab0 ) = bL(a0 ) = L(a0 ) loga0 (a). So, with a0 fixed and a allowed to vary, L(a) is just the logarithm function loga0 (a) scaled by L(a0 ). Looking at the graph of (a0 )x , it is geometrically clear that L(a0 ) is positive (though we have not proved that L(a0 ) is even defined). Thus the graph of L(a) looks qualitatively like the graph of loga0 (a). In particular, for a less than 1, L(a) is negative. The value L(1) equals 0. And L(a) approaches +∞ and a increases. Therefore, there must be a number where L takes the value 1. By long tradition, this number is called e; eh − 1 = 1. L(e) = lim h→0 h This is the definition of e. It sheds very little light on the decimal value of e. Because e is so important, the logarithm with base e is given a special name: the natural loga­ rithm. It is denote by, ln(a) = loge (a). So, finally, L(a) equals, L(a) = loge (a)L(e) = ln(a)(1) = ln(a). This leads to the formula for the derivative of ax , d(ax ) = ln(a)ax . dx In particular, d(ex ) = ex . dx In fact, ex is characterized by the property above and the property that e0 equals 1. 4. The derivative of loga (x) and the value of e. By the chain rule, d(au ) du = ln(a)au . dx dx For u = loga (x), au equals x. Thus, d(au ) d(x) = = 1. dx dx Thus, ln(a)au du = 1. dx 18.01 Calculus Jason Starr Fall 2005 Solving gives, d loga (x) 1 1 = = 1/(ln(a)x) . dx ln(a) au In particular, for a = e, this gives, d ln(x) = 1/x . dx What is the derivative of ln(x) at x = 1? On the one hand, since the derivative of ln(x) equals 1/x, the derivative at x = 1 is 1/1 = 1. On the other hand, the definition of the derivative gives, ln(1 + h) − ln(1) . h→0 h lim Of course, ln(1) equals 0, so this simplifies to, 1 ln(1 + h). h→0 h lim Using Rule 2 for logarithms, this gives, lim ln((1 + h)1/h ). h→0 Since ln(y) is continuous, the limit equals, ln[lim(1 + h)1/h ]. h→0 So the natural logarithm of the inner limit equals 1. But e is the unique number whose natural logarithm equals 1. This leads to the formula, e = lim(1 + h)1/h . h→0 Making the substitution n = 1/h leads to the more familiar form, lim (1 + 1/n)n = e . n→+∞ This can be used to compute e to arbitrary accuracy. 2.718281828459045... The first few digits of e are 5. Logarithmic differentiation. There is a method of computing derivatives of products of functions that is often useful. If y is a product of n factors, say f1 (x)· f2 (x)·· · ·· fn (x), the derivative of y can be computed by the product rule. However, it seems to be a fact that multiplication is more error­prone than addition. Thus introduce, u = ln(y) = ln(f1 (x)) + ln(f2 (x)) + · · · + ln(fn (x)). 18.01 Calculus Jason Starr Fall 2005 The derivative of u is, du d = (ln(f1 (x))) + · · · + dx dx Using the chain rule, this is, du f � (x) = 1 + ··· + dx f1 (x) d (ln(fn (x))). dx fn� (x) . fn (x) Thus, far fewer multiplications are needed to compute u� . This is good, because also, du d ln(y) 1 dy = = . dx dx y dx Therefore the derivative of y can be computed as, � � � y = yu = (f1 (x) · · · · · fn (x)) Example. Let y be, (1 + x3 )(1 + x3/7 f1� (x) f � (x) + ··· + n f1 (x) fn (x) √ x) � . . Then, √ 3 u = ln(y) = ln(1 + x3 ) + ln(1 + x) − ln(x). 7 √ � √ � √ √ 3 � 2 3 By the chain rule, ln(1+x ) = 3x /(1+x ) and ln(1+ x) = ( x) /(1+ x) = (1/2x−1/2 )/(1+ x). Thus, u� equals, 3x2 1 3 √ − . u� = + √ 3 (1 + x ) 2 x(1 + x) 7x So, finally, (1 + x3 )(1 + y = yu = x3/7 � � √ x) � 1 3 3x2 √ − + √ 3 (1 + x ) 2 x(1 + x) 7x � .