Lesson 14: Derivatives of Exponential and Logarithmic Functions (Section 021 slides)

Section 3.3 Derivatives of Exponential and Logarithmic Functions V63.0121.021, Calculus I New York University October 25, 2010 Announcements Midterm is graded. Please see FAQ. Quiz 3 next week on 2.6, 2.8, 3.1, 3.2

Announcements Midterm is graded. Please see FAQ. Quiz 3 next week on 2.6, 2.8, 3.1, 3.2 V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 2 / 38

Objectives Know the derivatives of the exponential functions (with any base) Know the derivatives of the logarithmic functions (with any base) Use the technique of logarithmic diﬀerentiation to ﬁnd derivatives of functions involving roducts, quotients, and/or exponentials. V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 3 / 38

Outline Recall Section 3.1–3.2 Derivative of the natural exponential function Exponential Growth Derivative of the natural logarithm function Derivatives of other exponentials and logarithms Other exponentials Other logarithms Logarithmic Diﬀerentiation The power rule for irrational powers V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 4 / 38

Conventions on power expressions Let a be a positive real number. If n is a positive whole number, then an = a · a · · · · · a n factors 0 a = 1. 1 For any real number r , a−r = . ar √ For any positive whole number n, a1/n = n a. There is only one continuous function which satisﬁes all of the above. We call it the exponential function with base a. V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 5 / 38

Properties of exponential Functions Theorem If a > 0 and a = 1, then f (x) = ax is a continuous function with domain (−∞, ∞) and range (0, ∞). In particular, ax > 0 for all x. For any real numbers x and y , and positive numbers a and b we have ax+y = ax ay ax ax−y = y a (ax )y = axy (ab)x = ax b x V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 6 / 38

Properties of exponential Functions Theorem If a > 0 and a = 1, then f (x) = ax is a continuous function with domain (−∞, ∞) and range (0, ∞). In particular, ax > 0 for all x. For any real numbers x and y , and positive numbers a and b we have ax+y = ax ay ax ax−y = y (negative exponents mean reciprocals) a (ax )y = axy (ab)x = ax b x V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 6 / 38

Properties of exponential Functions Theorem If a > 0 and a = 1, then f (x) = ax is a continuous function with domain (−∞, ∞) and range (0, ∞). In particular, ax > 0 for all x. For any real numbers x and y , and positive numbers a and b we have ax+y = ax ay ax ax−y = y (negative exponents mean reciprocals) a (ax )y = axy (fractional exponents mean roots) (ab)x = ax b x V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 6 / 38

Graphs of various exponential functions y y = ((21/2))xx (1/3)x y = /3 =y y = (1/10)xy = 10x= 3xy = 2x y y = 1.5x y = 1x x V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 7 / 38

The magic number Deﬁnition n 1 e = lim 1+ = lim+ (1 + h)1/h n→∞ n h→0 V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 8 / 38

Existence of e See Appendix B n 1 n 1+ We can experimentally verify n that this number exists and 1 2 is 2 2.25 3 2.37037 e ≈ 2.718281828459045 . . . 10 2.59374 100 2.70481 e is irrational 1000 2.71692 e is transcendental 106 2.71828 V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 9 / 38

Logarithms Deﬁnition The base a logarithm loga x is the inverse of the function ax y = loga x ⇐⇒ x = ay The natural logarithm ln x is the inverse of e x . So y = ln x ⇐⇒ x = e y . Facts (i) loga (x1 · x2 ) = loga x1 + loga x2 x1 (ii) loga = loga x1 − loga x2 x2 (iii) loga (x r ) = r loga x V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 10 / 38

Graphs of logarithmic functions y 10=yx= y = yy = 3e x 2x x y = log2 x y = ln x y = log3 x (0, 1) y = log10 x (1, 0) x V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 11 / 38

Change of base formula for logarithms Fact If a > 0 and a = 1, and the same for b, then logb x loga x = logb a Proof. If y = loga x, then x = ay So logb x = logb (ay ) = y logb a Therefore logb x y = loga x = logb a V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 12 / 38

Upshot of changing base The point of the change of base formula logb x 1 loga x = = · logb x = (constant) · logb x logb a logb a is that all the logarithmic functions are multiples of each other. So just pick one and call it your favorite. Engineers like the common logarithm log = log10 Computer scientists like the binary logarithm lg = log2 Mathematicians like natural logarithm ln = loge Naturally, we will follow the mathematicians. Just don’t pronounce it “lawn.” V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 13 / 38

Derivatives of Exponential Functions Fact If f (x) = ax , then f (x) = f (0)ax . V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 15 / 38

Derivatives of Exponential Functions Fact If f (x) = ax , then f (x) = f (0)ax . Proof. Follow your nose: f (x + h) − f (x) ax+h − ax f (x) = lim = lim h→0 h h→0 h a x ah − ax a h−1 = lim = ax · lim = ax · f (0). h→0 h h→0 h V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 15 / 38

Derivatives of Exponential Functions Fact If f (x) = ax , then f (x) = f (0)ax . Proof. Follow your nose: f (x + h) − f (x) ax+h − ax f (x) = lim = lim h→0 h h→0 h a x ah − ax a h−1 = lim = ax · lim = ax · f (0). h→0 h h→0 h To reiterate: the derivative of an exponential function is a constant times that function. Much diﬀerent from polynomials! V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 15 / 38

The funny limit in the case of e Remember the deﬁnition of e: n 1 e = lim 1+ = lim (1 + h)1/h n→∞ n h→0 Question eh − 1 What is lim ? h→0 h V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 16 / 38

The funny limit in the case of e Remember the deﬁnition of e: n 1 e = lim 1+ = lim (1 + h)1/h n→∞ n h→0 Question eh − 1 What is lim ? h→0 h Answer If h is small enough, e ≈ (1 + h)1/h . So h eh − 1 (1 + h)1/h −1 (1 + h) − 1 h ≈ = = =1 h h h h V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 16 / 38

The funny limit in the case of e Remember the deﬁnition of e: n 1 e = lim 1+ = lim (1 + h)1/h n→∞ n h→0 Question eh − 1 What is lim ? h→0 h Answer If h is small enough, e ≈ (1 + h)1/h . So h eh − 1 (1 + h)1/h −1 (1 + h) − 1 h ≈ = = =1 h h h h eh − 1 So in the limit we get equality: lim =1 h→0 h V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 16 / 38

Derivative of the natural exponential function From d x ah − 1 eh − 1 a = lim ax and lim =1 dx h→0 h h→0 h we get: Theorem d x e = ex dx V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 17 / 38

Exponential Growth Commonly misused term to say something grows exponentially It means the rate of change (derivative) is proportional to the current value Examples: Natural population growth, compounded interest, social networks V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 18 / 38

Examples Examples Find derivatives of these functions: e 3x 2 ex x 2e x V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 19 / 38

Examples Examples Find derivatives of these functions: e 3x 2 ex x 2e x Solution d 3x e = 3e 3x dx V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 19 / 38

Examples Examples Find derivatives of these functions: e 3x 2 ex x 2e x Solution d 3x e = 3e 3x dx d x2 2 d 2 e = ex (x 2 ) = 2xe x dx dx V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 19 / 38

Examples Examples Find derivatives of these functions: e 3x 2 ex x 2e x Solution d 3x e = 3e 3x dx d x2 2 d 2 e = ex (x 2 ) = 2xe x dx dx d 2 x x e = 2xe x + x 2 e x dx V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 19 / 38

Derivative of the natural logarithm function Let y = ln x. Then x = e y so V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 21 / 38

Derivative of the natural logarithm function Let y = ln x. Then x = e y so dy ey =1 dx V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 21 / 38

Derivative of the natural logarithm function Let y = ln x. Then x = e y so dy ey =1 dx dy 1 1 =⇒ = y = dx e x V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 21 / 38

Derivative of the natural logarithm function Let y = ln x. Then x = e y so dy ey =1 dx dy 1 1 =⇒ = y = dx e x We have discovered: Fact d 1 ln x = dx x V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 21 / 38

Derivative of the natural logarithm function y Let y = ln x. Then x = e y so dy ey =1 dx ln x dy 1 1 =⇒ = y = dx e x x We have discovered: Fact d 1 ln x = dx x V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 21 / 38

Derivative of the natural logarithm function y Let y = ln x. Then x = e y so dy ey =1 dx ln x dy 1 1 1 =⇒ = y = x dx e x x We have discovered: Fact d 1 ln x = dx x V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 21 / 38

The Tower of Powers y y x3 3x 2 The derivative of a power x2 2x 1 function is a power function x1 1x 0 of one lower power x0 0 ? ? x −1 −1x −2 x −2 −2x −3 V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 22 / 38

The Tower of Powers y y x3 3x 2 The derivative of a power x2 2x 1 function is a power function x1 1x 0 of one lower power Each power function is the x0 0 derivative of another power ? x −1 function, except x −1 x −1 −1x −2 x −2 −2x −3 V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 22 / 38

The Tower of Powers y y x3 3x 2 The derivative of a power x2 2x 1 function is a power function x1 1x 0 of one lower power Each power function is the x0 0 derivative of another power ln x x −1 function, except x −1 x −1 −1x −2 ln x ﬁlls in this gap precisely. x −2 −2x −3 V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 22 / 38

Examples Examples Find derivatives of these functions: ln(3x) x ln x √ ln x V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 23 / 38

Examples Example d Find ln(3x). dx V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 24 / 38

Examples Example d Find ln(3x). dx Solution (chain rule way) d 1 1 ln(3x) = ·3= dx 3x x V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 24 / 38

Examples Example d Find ln(3x). dx Solution (chain rule way) d 1 1 ln(3x) = ·3= dx 3x x Solution (properties of logarithms way) d d 1 1 ln(3x) = (ln(3) + ln(x)) = 0 + = dx dx x x V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 24 / 38

Examples Example d Find ln(3x). dx Solution (chain rule way) d 1 1 ln(3x) = ·3= dx 3x x Solution (properties of logarithms way) d d 1 1 ln(3x) = (ln(3) + ln(x)) = 0 + = dx dx x x The ﬁrst answer might be surprising until you see the second solution. V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 24 / 38

Examples Example d Find x ln x dx V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 25 / 38

Examples Example d Find x ln x dx Solution The product rule is in play here: d d d 1 x ln x = x ln x + x ln x = 1 · ln x + x · = ln x + 1 dx dx dx x V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 25 / 38

Examples Example d √ Find ln x. dx V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 26 / 38

Examples Example d √ Find ln x. dx Solution (chain rule way) d √ 1 d √ 1 1 1 ln x = √ x=√ √ = dx x dx x2 x 2x V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 26 / 38

Examples Example d √ Find ln x. dx Solution (chain rule way) d √ 1 d √ 1 1 1 ln x = √ x=√ √ = dx x dx x2 x 2x Solution (properties of logarithms way) d √ d 1 1 d 1 1 ln x = ln x = ln x = · dx dx 2 2 dx 2 x V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 26 / 38

Examples Example d √ Find ln x. dx Solution (chain rule way) d √ 1 d √ 1 1 1 ln x = √ x=√ √ = dx x dx x2 x 2x Solution (properties of logarithms way) d √ d 1 1 d 1 1 ln x = ln x = ln x = · dx dx 2 2 dx 2 x The ﬁrst answer might be surprising until you see the second solution. V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 26 / 38

Other logarithms Example d x Use implicit diﬀerentiation to ﬁnd a . dx V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 28 / 38

Other logarithms Example d x Use implicit diﬀerentiation to ﬁnd a . dx Solution Let y = ax , so ln y = ln ax = x ln a V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 28 / 38

Other logarithms Example d x Use implicit differentiation to find a . dx Solution Let y = ax , so ln y = ln ax = x ln a Differentiate implicitly: 1 dy dy = ln a =⇒ = (ln a)y = (ln a)ax y dx dx V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 28 / 38

Other logarithms Example d x Use implicit differentiation to find a . dx Solution Let y = ax , so ln y = ln ax = x ln a Differentiate implicitly: 1 dy dy = ln a =⇒ = (ln a)y = (ln a)ax y dx dx Before we showed y = y (0)y , so now we know that ah − 1 ln a = lim h→0 h V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 28 / 38

Other logarithms Example d Find loga x. dx V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 29 / 38

Other logarithms Example d Find loga x. dx Solution Let y = loga x, so ay = x. V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 29 / 38

Other logarithms Example d Find loga x. dx Solution Let y = loga x, so ay = x. Now diﬀerentiate implicitly: dy dy 1 1 (ln a)ay = 1 =⇒ = y = dx dx a ln a x ln a V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 29 / 38

Other logarithms Example d Find loga x. dx Solution Let y = loga x, so ay = x. Now diﬀerentiate implicitly: dy dy 1 1 (ln a)ay = 1 =⇒ = y = dx dx a ln a x ln a Another way to see this is to take the natural logarithm: ln x ay = x =⇒ y ln a = ln x =⇒ y = ln a dy 1 1 So = . dx ln a x V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 29 / 38

More examples Example d Find log2 (x 2 + 1) dx V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 30 / 38

More examples Example d Find log2 (x 2 + 1) dx Answer dy 1 1 2x = 2+1 (2x) = dx ln 2 x (ln 2)(x 2 + 1) V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 30 / 38

A nasty derivative Example √ (x 2 + 1) x + 3 Let y = . Find y . x −1 V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 32 / 38

A nasty derivative Example √ (x 2 + 1) x + 3 Let y = . Find y . x −1 Solution We use the quotient rule, and the product rule in the numerator: √ √ (x − 1) 2x x + 3 + (x 2 + 1) 1 (x + 3)−1/2 − (x 2 + 1) x + 3(1) 2 y = (x − 1)2 √ 2 + 1) √ 2x x + 3 (x (x 2 + 1) x + 3 = + √ − (x − 1) 2 x + 3(x − 1) (x − 1)2 V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 32 / 38

Another way √ (x 2 + 1) x + 3 y = x −1 1 ln y = ln(x 2 + 1) + ln(x + 3) − ln(x − 1) 2 1 dy 2x 1 1 = 2 + − y dx x + 1 2(x + 3) x − 1 So dy 2x 1 1 = + − y dx x2 + 1 2(x + 3) x − 1 √ 2x 1 1 (x 2 + 1) x + 3 = 2+1 + − x 2(x + 3) x − 1 x −1 V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 33 / 38

Compare and contrast Using the product, quotient, and power rules: √ √ 2x x + 3 (x 2 + 1) (x 2 + 1) x + 3 y = + √ − (x − 1) 2 x + 3(x − 1) (x − 1)2 Using logarithmic diﬀerentiation: √ 2x 1 1 (x 2 + 1) x + 3 y = 2+1 + − x 2(x + 3) x − 1 (x − 1) V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38

Compare and contrast Using the product, quotient, and power rules: √ √ 2x x + 3 (x 2 + 1) (x 2 + 1) x + 3 y = + √ − (x − 1) 2 x + 3(x − 1) (x − 1)2 Using logarithmic diﬀerentiation: √ 2x 1 1 (x 2 + 1) x + 3 y = 2+1 + − x 2(x + 3) x − 1 (x − 1) Are these the same? V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38

Compare and contrast Using the product, quotient, and power rules: √ √ 2x x + 3 (x 2 + 1) (x 2 + 1) x + 3 y = + √ − (x − 1) 2 x + 3(x − 1) (x − 1)2 Using logarithmic diﬀerentiation: √ 2x 1 1 (x 2 + 1) x + 3 y = 2+1 + − x 2(x + 3) x − 1 (x − 1) Are these the same? Yes. V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38

Compare and contrast Using the product, quotient, and power rules: √ √ 2x x + 3 (x 2 + 1) (x 2 + 1) x + 3 y = + √ − (x − 1) 2 x + 3(x − 1) (x − 1)2 Using logarithmic diﬀerentiation: √ 2x 1 1 (x 2 + 1) x + 3 y = 2+1 + − x 2(x + 3) x − 1 (x − 1) Are these the same? Yes. Which do you like better? V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38

Compare and contrast Using the product, quotient, and power rules: √ √ 2x x + 3 (x 2 + 1) (x 2 + 1) x + 3 y = + √ − (x − 1) 2 x + 3(x − 1) (x − 1)2 Using logarithmic diﬀerentiation: √ 2x 1 1 (x 2 + 1) x + 3 y = 2+1 + − x 2(x + 3) x − 1 (x − 1) Are these the same? Yes. Which do you like better? What kinds of expressions are well-suited for logarithmic diﬀerentiation? V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38

Compare and contrast Using the product, quotient, and power rules: √ √ 2x x + 3 (x 2 + 1) (x 2 + 1) x + 3 y = + √ − (x − 1) 2 x + 3(x − 1) (x − 1)2 Using logarithmic diﬀerentiation: √ 2x 1 1 (x 2 + 1) x + 3 y = 2+1 + − x 2(x + 3) x − 1 (x − 1) Are these the same? Yes. Which do you like better? What kinds of expressions are well-suited for logarithmic diﬀerentiation? Products, quotients, and powers V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38

Derivatives of powers y Question Let y = x x . Which of these is true? (A) Since y is a power function, y = x · x x−1 = x x . (B) Since y is an exponential 1 function, y = (ln x) · x x (C) Neither x 1 V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 35 / 38

Derivatives of powers y Question Let y = x x . Which of these is true? (A) Since y is a power function, y = x · x x−1 = x x . (B) Since y is an exponential 1 function, y = (ln x) · x x (C) Neither x 1 Answer (A) This can’t be y because x x > 0 for all x > 0, and this function decreases at some places (B) This can’t be y because (ln x)x x = 0 when x = 1, and this function does not have a horizontal tangent at x = 1. V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 35 / 38

It’s neither! Or both? Solution If y = x x , then ln y = x ln x 1 dy 1 = x · + ln x = 1 + ln x y dx x dy = (1 + ln x)x x = x x + (ln x)x x dx V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 36 / 38

It’s neither! Or both? Solution If y = x x , then ln y = x ln x 1 dy 1 = x · + ln x = 1 + ln x y dx x dy = (1 + ln x)x x = x x + (ln x)x x dx Remarks Each of these terms is one of the wrong answers! V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 36 / 38

It’s neither! Or both? Solution If y = x x , then ln y = x ln x 1 dy 1 = x · + ln x = 1 + ln x y dx x dy = (1 + ln x)x x = x x + (ln x)x x dx Remarks Each of these terms is one of the wrong answers! y < 0 on the interval (0, e −1 ) y = 0 when x = e −1 V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 36 / 38

Derivatives of power functions with any exponent Fact (The power rule) Let y = x r . Then y = rx r −1 . V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 37 / 38

Derivatives of power functions with any exponent Fact (The power rule) Let y = x r . Then y = rx r −1 . Proof. y = x r =⇒ ln y = r ln x Now diﬀerentiate: 1 dy r = y dx x dy y =⇒ = r = rx r −1 dx x V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 37 / 38

Summary Derivatives of logarithmic and exponential functions: y y ex ex ax (ln a) · ax 1 ln x x 1 1 loga x · ln a x Logarithmic Diﬀerentiation can allow us to avoid the product and quotient rules. We are ﬁnally done with the Power Rule! V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 38 / 38

Lesson 14: Derivatives of Exponential and Logarithmic Functions (Section 021 slides)

by Matthew Leingang

Accessibility

Categories

Upload Details

Usage Rights

Statistics

Lesson 14: Derivatives of Exponential and Logarithmic Functions (Section 021 slides) Presentation Transcript