UNIT I - THE DERIVATIVES Module 7: Differentiation of Logarithmic and Exponential Functions Introduction In this topic, we will discuss differentiation of the other set of transcendental functions, which are logarithmic and exponential functions Learning Objectives At the end of this topic, you should be able to demonstrate the following: a. Determine the derivative of logarithmic functions; b. Find the derivative of exponential functions; and c. Differentiate a function raised to a function by logarithmic differentiation. Presentation of Contents Differentiation of Logarithmic Functions Let u be a differentiable function of x, that is, u = g(x). The following formulas are used in differentiating logarithmic functions. ๐ (๐๐๐๐ ๐) ๐ ๐ 1. ๐ ๐ = ๐๐๐๐ ๐ . ๐ ๐ ๐ ๐ If b = 10, ๐กโ๐๐ ๐๐๐๐ ๐ข = ๐๐๐๐ข called the common logarithm. The formula is reduced to ๐ (๐๐๐๐) ๐ ๐ ๐ ๐ = log๐ . ๐ ๐ ๐ ๐ If b = e, ๐กโ๐๐ ๐๐๐๐ ๐ข = ๐๐๐ข called the natural logarithm. The formula is reduced to ๐ ๐ 2. ๐ ๐ (๐ฅ๐ง ๐) = ๐ . ๐ ๐ ๐ ๐ EXAMPLE 1 ๐๐ฆ a. Find ๐๐ฅ if y = ๐๐๐5 (3๐ฅ 2 − 5). ๐๐ฆ y = ๐๐๐5 (3๐ฅ 2 − 5) takes the form of y = ๐๐๐๐ ๐ข with b = 5 and u = 3๐ฅ 2 − 5. To find ๐๐ฅ , we use the formula ๐๐ฆ ๐๐ฅ ๐๐ฆ ๐๐ฅ ๐๐ฆ ๐๐ฅ ๐๐ฆ ๐๐ฅ = = = = ๐ ๐๐ฅ 1 ๐๐ข (๐๐๐๐ ๐ข) = ๐ข ๐๐๐๐ ๐ . ๐๐ฅ 1 3๐ฅ2 −5 1 3๐ฅ2 −5 ๐ 2 ๐๐๐5 ๐ . ๐๐ฅ (3๐ฅ − 5) ๐๐๐5 ๐ . (6x) 6๐ฅ๐๐๐5 ๐ 3๐ฅ2 −5 by substitution b. Find ๐๐ฆ ๐๐ฅ if y = log ๐ ๐๐2 4๐ฅ. ๐๐ฆ Again, y = log ๐ ๐๐2 4๐ฅ takes the form of y = log ๐ข with u =๐ ๐๐2 4๐ฅ. To find ๐๐ฅ , we use the formula ๐ (๐๐๐๐ข) ๐๐ฅ ๐๐ฆ ๐๐ฅ ๐๐ฆ ๐๐ฅ ๐๐ฆ ๐๐ฅ 1 ๐ข = log๐ . 1 ๐๐ข ๐๐ฅ by substitution ๐ = ๐ ๐๐2 4๐ฅ log๐ . ๐๐ฅ (๐ ๐๐2 4๐ฅ) log๐ ๐ = ๐ ๐๐2 4๐ฅ . ๐๐ฅ [(sin 4๐ฅ)2 ] log๐ by GPR ๐ = ๐ ๐๐2 4๐ฅ . 2 (sin 4๐ฅ)2−1 ๐๐ฅ (sin 4๐ฅ) ๐๐ฆ log๐ ๐ = ๐ ๐๐2 4๐ฅ . 2 sin 4x [cos 4๐ฅ . ๐๐ฅ (4๐ฅ) ] ๐๐ฅ ๐๐ฆ ๐๐ฅ ๐๐ฆ ๐๐ฅ ๐๐ฆ log๐ = ๐ ๐๐2 4๐ฅ . 2 sin 4x [cos 4๐ฅ . (4)] = = ๐๐ฅ 8log๐ sin 4๐ฅ cos 4๐ฅ ๐ ๐๐2 4๐ฅ 8log๐๐๐๐ 4๐ฅ = 8log๐ cot 4๐ฅ sin 4๐ฅ ๐๐ฆ c. Find ๐๐ฅ if y = ln( 4x + 3). ๐๐ฆ y = ln( 4x + 3) takes the form of y = ln u with u = 4x + 3. To find ๐๐ฅ , we use the formula ๐ ๐๐ฅ ๐๐ฆ 1 ๐ข (ln ๐ข) = . 1 = 4x + 3. ๐๐ฅ ๐๐ฆ ๐๐ฅ ๐๐ฆ ๐๐ข ๐๐ฅ ๐ ๐๐ฅ (4๐ฅ + 3) 1 = 4x + 3. (4) 4 = 4x + 3 ๐๐ฅ d. Find f ‘(x) if f(x) = ๐๐4 (๐ฅ + 3) f(x) = ๐๐4 (๐ฅ + 3) can be expressed in power form as f(x) = [ln(๐ฅ + 3)]4 Note: ๐๐๐ (๐ + ๐) ≠ ln(๐ + ๐)๐ f(x) = [ln(๐ฅ + 3)]4 takes the form of f(x) = ๐ข๐ where u = ln(๐ฅ + 3) and x = 4. To find f ‘(x), we use the GPR. ๐๐ข f’(x) = n๐ข๐−1 . ๐๐ฅ u = ln(๐ฅ + 3) ๐๐ข ๐๐ฅ 1 ๐ = ๐ฅ+3 ๐๐ฅ (๐ฅ + 3) n=4 ๐๐ข 1 = ๐ฅ+3 (1) ๐๐ฅ ๐๐ข ๐๐ฅ 1 = ๐ฅ+3 1 f ‘(x) = 4 [ln(๐ฅ + 3)]3 ๐ฅ+3 f ‘(x) = 4๐๐3 (x+3) ๐ฅ+3 ๐ฅ2 ๐๐ฆ e. Find ๐๐ฅ given that y = ๐๐๐ฅ ๐ฅ2 ๐๐ฆ y = ๐๐๐ฅ takes the form of a quotient of two functions, so to find for ๐๐ฅ , we use the quotient rule. ๐๐ฆ = ๐๐ฅ ๐(๐ฅ)๐′ (๐ฅ)−๐(๐ฅ)๐′(๐ฅ) [๐(๐ฅ)]2 ๐ฅ2 y = ๐๐๐ฅ n(x) = ๐ฅ 2 d(x) = lnx n’(x) = 2x d’(x) = ๐ฅ ๐๐ฅ (๐ฅ) 1 ๐ 1 d’(x) = ๐ฅ ๐๐ฆ = ๐๐ฅ ๐๐ฆ ๐๐ฅ ๐๐ฆ ๐๐ฅ = = 1 ๐ฅ ๐๐๐ฅ(2๐ฅ)−๐ฅ 2 ( ) (๐๐๐ฅ)2 2๐ฅ๐๐๐ฅ−๐ฅ ๐๐2 ๐ฅ or ๐ฅ(2๐๐๐ฅ−1) ๐๐2 ๐ฅ Application Activity 1 ๐๐ฆ Find the ๐๐ฅ of each function below. 3 1. y = log √√12๐ฅ 2. y = ๐๐2 ๐ฅ − ln(๐๐๐ฅ 2 ) 3. y = ln(sec ๐ฅ + tan ๐ฅ) 4. y = ln ๐ฅ arctan x 5. xlny = x + ylnx Differentiation of Exponential Functions Let u be a differentiable function of x, that is, u = g(x). The following formulas are used in differentiating exponential functions. 1. 2. ๐ ๐ ๐ ๐ ๐ ๐ (๐๐ ) = ๐๐ ln a . (๐๐ ) = ๐๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ Where: a = constant and a > 0 and a ≠1 e = 2.718281828… ( e is an irrational number) EXAMPLE 2 ๐๐ฆ a. Find the ๐๐ฅ if y = 34๐ฅ ๐๐ฆ y = 34๐ฅ is in the form of y = ๐๐ข . To find ๐๐ฅ , we use the formula ๐๐ฆ ๐๐ฅ ๐ ๐ ๐ = ๐ ๐ (๐๐ ) = ๐๐ ln a . ๐ ๐ Where a = 3, ๐๐ฆ ๐๐ฅ ๐๐ฆ ๐๐ฅ ๐๐ฆ ๐๐ฅ u = 4x ๐ = 34๐ฅ ln 3 . ๐ ๐ (4x) = 34๐ฅ ln 3 . 4 = 4(34๐ฅ )ln3 ๐๐ฆ b. Given y = ๐ 2๐ฅ+3 , find ๐๐ฅ . ๐๐ฆ Again, y = ๐ 2๐ฅ+3 is in the form of y = ๐ ๐ข . To find ๐๐ฅ , we use the formula ๐๐ฆ ๐๐ฅ ๐ = ๐ ๐ (๐๐ ) = ๐๐ ๐ ๐ ๐ ๐ u = 2x + 3 ๐๐ฆ ๐๐ฅ ๐๐ฆ ๐๐ฅ ๐๐ฆ ๐๐ฅ ๐ = ๐ 2๐ฅ+3 ๐๐ฅ (2x + 3) = ๐ 2๐ฅ+3 (2) = 2 ๐ 2๐ฅ+3 c. Find f ‘(x) if f(x) = ๐ ๐ฅ ( ๐ 2๐ฅ - ๐ ๐ฅ ) f(x) = ๐ ๐ฅ ( ๐ 2๐ฅ - ๐ ๐ฅ ) is a product of two functions. Now, instead of using the product rule, simplify the function using DPMA. That is, f(x) = ๐ ๐ฅ ( ๐ 2๐ฅ - ๐ ๐ฅ ) using the DPMA f(x) = ๐ ๐ฅ ๐ 2๐ฅ - ๐ ๐ฅ ๐ ๐ฅ using the product law of exponents f(x) = ๐ ๐ฅ+ 2๐ฅ - ๐ ๐ฅ+๐ฅ simplifying the exponents f(x) = ๐ 3๐ฅ - ๐ 2๐ฅ Find f ‘(x) using the sum rule ๐ ๐ f ’(x) = ๐๐ฅ (๐ 3๐ฅ ) - ๐๐ฅ (๐ 2๐ฅ ) ๐ ๐ f ‘(x) = ๐ 3๐ฅ . ๐๐ฅ (3x) - ๐ 2๐ฅ . ๐๐ฅ (2x) f ‘(x) = ๐ 3๐ฅ (3) - ๐ 2๐ฅ (2) f ‘(x) = 3๐ 3๐ฅ - 2๐ 2๐ฅ ๐๐ฆ d. Given y = √1 + ๐ ๐ฅ , find ๐๐ฅ . 1 y = √1 + ๐ ๐ฅ in power form is ๐ฆ = (1 + ๐ ๐ฅ )2 . Since the function is in power form y = ๐ข๐ , ๐๐ฆ we use the GPR to find the ๐๐ฅ . ๐๐ฆ ๐๐ฅ ๐๐ข = n๐ข๐−1 . ๐๐ฅ 1 ๐ฆ = (1 + ๐ ๐ฅ )2 ๐๐ฆ ๐๐ฅ ๐๐ฆ ๐๐ฅ ๐๐ฆ ๐๐ฅ ๐๐ฆ ๐๐ฅ ๐๐ฆ ๐๐ฅ u = 1 + ๐๐ฅ 1 n=2 ๐๐ข = n๐ข๐−1 . ๐๐ฅ = = = = 1 −1 1 −1 1 −1 ๐ (1 + ๐ ๐ฅ ) 2 . (1 + ๐ ๐ฅ ) 2 ๐๐ฅ ๐ (1 + ๐ ๐ฅ ) 2 . (0 + ๐ ๐ฅ (๐ฅ)) 2 ๐๐ฅ (1 + ๐ ๐ฅ ) 2 . ๐ ๐ฅ 2 ๐๐ฅ 2 √1+ ๐ ๐ฅ ๐ 3๐ฅ +1 e. Find f ‘(x) if f(x) = ๐ 3๐ฅ −1 ๐ 3๐ฅ +1 f(x) = ๐ 3๐ฅ −1 is in the form of a quotient, so we use the quotient rule. f ‘(x) = ๐(๐ฅ)๐′ (๐ฅ)−๐(๐ฅ)๐′(๐ฅ) [๐(๐ฅ)]2 n(x) = ๐ 3๐ฅ + 1 n’(x) = ๐ 3๐ฅ f ‘(x) = ๐ ๐๐ฅ (3๐ฅ) + 0 d(x) = ๐ 3๐ฅ − 1 d’(x) = ๐ 3๐ฅ ๐ ๐๐ฅ (3๐ฅ) – 0 n’(x) = ๐ 3๐ฅ (3) d’(x) = ๐ 3๐ฅ (3) n’(x) = 3๐ 3๐ฅ d’(x) = 3๐ 3๐ฅ (๐ 3๐ฅ −1)3๐ 3๐ฅ −(๐ 3๐ฅ +1)3๐ 3๐ฅ [๐3๐ฅ −1] 2 Using DPMA in the numerator f ‘(x) = f ‘(x) = 3๐ 6๐ฅ − 3๐ 3๐ฅ −3๐ 6๐ฅ − 3๐ 3๐ฅ [๐3๐ฅ −1] 2 combining similar terms in the numerator − 6๐ 3๐ฅ [๐3๐ฅ −1] 2 Application Activity 2 ๐๐ฆ Find the ๐๐ฅ for each function below. 3 1. y = √(๐ −2๐ฅ + 2๐ ๐ฅ )2 2. y = 4๐ฅ ln 4๐ฅ 3. y = (1 − ๐ −๐ฅ )2 ( 1 + ๐ ๐ฅ ) 4. y = arctan√๐ ๐ฅ − 1 ๐ 2๐ฅ −1 5. y = ln (๐ 2๐ฅ +1) Before we move to the last differentiation technique, observe these three groups of functions. y = ๐ฅ4 y = 4๐ฅ y = ๐ฅ๐ฅ y = (2๐ฅ + 1)๐ y = ๐ (2๐ฅ+1) ๐ฆ = (2๐ฅ + 1)๐ y = ๐ฅ2 y = 2๐ฅ y = ๐ฅ2 ๐ฆ = (ln ๐ฅ)2 ๐ฆ = 2๐๐๐ฅ y = (ln ๐ฅ)2๐ฅ Power Functionsthe bases are functions of x and the exponents are constants Exponential Functions – The bases are constants and the exponents are functions of x Differentiate using power rule or GPR Differentiate using Differentiation of Exponential functions ๐ฅ ๐ฅ Power Functions whose base and exponents are both differentiable functions Differentiate using Logarithmic differentiation Note: Be able to recognize and differentiate these three types of functions and apply appropriate differentiation rules or techniques. Logarithmic Differentiation Consider a function y = ๐ข๐ฃ where u = g(x) and v = h(x). Then y = ๐(๐ฅ)โ(๐ฅ) is a power function whose base and exponent are both differentiable functions of x. To find the derivative of this kind of function, we may use the method of logarithmic differentiation defined by: If y = y = ๐ข๐ฃ where u = g(x) and v = h(x), then ๐ ๐ ๐ ๐ ๐ = ๐๐ ๐ ๐ (๐๐๐๐) EXAMPLE 2 ๐ ๐ a. Find ๐ ๐ if y = ๐๐ y = ๐๐ is in the form of y = ๐ข๐ฃ where u = x and v = x. To find ๐ ๐ , ๐ ๐ we use the logarithmic differentiation. That is, ๐ ๐ ๐ ๐ = ๐๐ ๐ (๐๐๐๐) ๐ ๐ ๐ ๐ ๐ ๐ = ๐๐ ๐ ๐ (๐๐๐๐) dy dx = x x [x (x . 1) + ln x (1)] dy dx = x x [1 + ln x ] ๐ ๐ using product rule to evaluate ๐๐ฅ (๐ฅ๐๐๐ฅ) 1 ๐ ๐ b. Find ๐ ๐ if ๐ฆ = (2๐ฅ + 1)๐ ๐ฅ ๐ฅ Again, ๐ฆ = (2๐ฅ + 1)๐ takes the form of y = ๐ข๐ฃ where u = 2x + 1 and v = ๐ ๐ฅ . ๐ ๐ ๐ ๐ = ๐๐ ๐ ๐ (๐๐๐๐) ๐ dy dx = (2๐ฅ + 1)๐ dy dx = (2๐ฅ + 1)๐ [๐ ๐ฅ ( dy dx = (2๐ฅ + 1)๐ [ ๐ฅ ๐ ๐ ๐ (๐ ๐ฅ ln(2x + 1)) 1 ๐ฅ 2๐ฅ+1 2๐ ๐ฅ ๐ฅ 2๐ฅ+1 using product rule to evaluate ๐ (๐๐ฅ ln(2x + 1)) ๐๐ฅ . 2) + ln(2x + 1)(๐ ๐ฅ . 1) ] + ๐ ๐ฅ ln(2x + 1) ] ๐๐ฆ c. Find ๐๐ฅ if ๐ฅ ๐ฆ + ๐ ๐ฅ = 8 ๐๐ฆ ๐ฅ ๐ฆ + ๐ ๐ฅ = 8 is an implicit function, so we use implicit differentiation to find for ๐๐ฅ . Differentiating both sides with respect to x, we have ๐ ๐๐ฅ ๐ ๐๐ฅ ๐ (๐ฅ ๐ฆ + ๐ ๐ฅ ) = ๐๐ฅ(8) ๐ (๐ฅ ๐ฆ ) + ๐๐ฅ (๐ ๐ฅ ) = 0 ๐ ๐ฅ ๐ฆ ๐๐ฅ(y ln x) + ๐ ๐ฅ (1) = 0 1 ๐ ๐ฅ ๐ฆ [๐ฆ (๐ฅ . 1) + ln ๐ฅ . ๐๐ฅ (๐ฆ)] + ๐ ๐ฅ = 0 ๐ฆ ๐ฅ ๐ฆ (๐ฅ + ๐๐ ๐ฅ ๐ฆ(๐ฅ ๐ฆ ) ๐ฅ ๐๐ฆ ๐๐ฅ )+ ๐ ๐ฅ = 0 ๐๐ฆ + ๐ฅ ๐ฆ ln x ๐๐ฅ + ๐ ๐ฅ = 0 ๐๐ฆ ๐ฅ ๐ฆ ln x ๐๐ฅ = -๐ ๐ฅ ๐๐ฆ = ๐๐ฅ ๐๐ฆ ๐๐ฅ = −๐ ๐ฅ − ๐ฅ ๐ฅ ๐ฆ ln x −๐ฅ๐๐ฅ − ๐ฆ(๐ฅ๐ฆ ) ๐ฅ ๐ฅ ๐ฆ ln x −๐ฅ๐ ๐ฅ − ๐ฆ(๐ฅ ๐ฆ ) (๐ฅ)๐ฅ ๐ฆ ln x ๐๐ฆ = ๐๐ฅ −๐ฅ๐ ๐ฅ − ๐ฆ(๐ฅ ๐ฆ ) (๐ฅ)๐ฅ ๐ฆ ln x ๐๐ฆ −๐ฅ๐ ๐ฅ − ๐ฆ(๐ฅ ๐ฆ ) = ๐๐ฅ ๐ฅ ๐ฆ(๐ฅ๐ฆ ) ๐๐ฆ = ๐๐ฅ ๐ฆ(๐ฅ ๐ฆ ) ๐ฅ ๐ฆ+1 ln x Application Activity 3 Find ๐๐ฆ ๐๐ฅ for each function below. 1. y = (ln ๐ฅ)2๐ฅ 2. y = (sin ๐ฅ)๐ฅ 3. y = ๐ฅ 2 ๐ฅ 4. y = ๐ฅ ln( 1+ ๐ 5. y = ๐ฅ √๐ฅ ๐ฅ)