Logarithmic Differentiation

6 Key excerpts on "Logarithmic Differentiation"

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  eBook - PDF

  The Calculus Lifesaver

  All the Tools You Need to Excel at Calculus

  • Adrian Banner(Author)
  • 2009(Publication Date)
  • Princeton University Press

    (Publisher)

  9.5 Logarithmic Differentiation Logarithmic Differentiation is a useful technique for dealing with derivatives of things like f ( x ) g ( x ) , where both the base and the exponent are functions of x . After all, how on earth would you find d dx ( x sin( x ) ) with what we have seen already? It doesn’t fit any of the rules. Still, we have these nice log rules which cut exponents down to size. If we let y = x sin( x ) , then ln( y ) = ln( x sin( x ) ) = sin( x ) ln( x ) by log rule #5 from Section 9.1.4 above. Now let’s differentiate both sides (implicitly) with respect to x : d dx (ln( y )) = d dx (sin( x ) ln( x )) . Let’s look at the right-hand side first. This is just a function of x and re-quires the product rule; you should check that the derivative works out to be cos( x ) ln( x ) + sin( x ) /x . Now let’s look at the left-hand side. To differentiate ln( y ) with respect to x (not y !), we should use the chain rule. Set u = ln( y ), so that du/dy = 1 /y . We need to find du/dx ; by the chain rule, du dx = du dy dy dx = 1 y dy dx . So, implicitly differentiating the equation ln( y ) = sin( x ) ln( x ) produces 1 y dy dx = cos( x ) ln( x ) + sin( x ) x . Now we just have to multiply both sides by y and then replace y by x sin( x ) : dy dx = cos( x ) ln( x ) + sin( x ) x y = cos( x ) ln( x ) + sin( x ) x x sin( x ) . 190 • Exponentials and Logarithms That’s the answer we’re looking for. (By the way, there is another way we could have done this problem. Instead of using the variable y , we could just have used our formula A = e ln( A ) to write x sin( x ) = e ln( x sin( x ) ) = e sin( x ) ln( x ) . Now I leave it to you to differentiate the right-hand side of this with respect to x by using the product and chain rules. When you’ve finished, you should replace e sin( x ) ln( x ) by x sin( x ) and check that you get the same answer as the original one above.) Let’s review the main technique.

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  eBook - ePub

  Mathematics for Engineers and Scientists

  • Alan Jeffrey(Author)
  • 2004(Publication Date)
  • Chapman and Hall/CRC

    (Publisher)

  The process is called Logarithmic Differentiation, though it should more properly be called differentiation by means of the logarithmic function. It is best illustrated by example. Suppose we need to find d y/ d x, where y = f (x) g (x) h (x). Taking natural logarithms and using their properties gives ln y = ln f (x) − ln g (x) − ln h (x). Differentiation with respect to x then. gives 1 y d y d x = f ′ (x) f (x) − g ′ (x) g (x) − h ′ (x) h (x), which when multiplied by y gives d y/ d x. The simplification arises from the fact that in general the expression for (1 /y) (d y /d x) is easier to calculate than d y /d x. Example 6.7 Find d y/ d x by means of Logarithmic Differentiation. if y = (2 x − 7) 5 (3 x + 1) 1 / 2 (2 x − 1) 3 / 2. Solution Taking the natural logarithm gives ln y = 5 ln (2 x − 7) − 1 2 ln (3 x + 1) − 3 2 ln (2 x − 1). Differentiating with respect to x we. find 1 y d y d x = 10 2 x − 7 − 3 2 (3 x + 1) − 3 2 x − 1, which after multiplication by y and simplification becomes d y d x = (72 x 2 + 142 x + 1) (2 x − 7) 4 2 (3 x + 1) 3 / 2 (2 x − 1) 5 / 2. 6.4  Hyperbolic functions It is useful to define new functions called the hyperbolic. sine, written sinhx and pronounced either as ‘cinch x’ or ‘shine x ’, and the hyperbolic cosine, written cosh x and pronounced ‘kosh x ’, which are related to the exponential function. This is achieved as follows. Definition 6.3 For all real x we defme sinh x and cosh x by the requirement that (hyperbolic functions) sinh x = e x − e − x 2, cosh x = e x + e − x 2. It is an immediate consequence of combining the series for. e x and e − x that sinh x = x + x 3 3 ! + x 5 5 ! + x 7 7 ! + … + x 2 n + 1 (2 n + 1) ! + … = ∑ n = 0 ∞ x 2 n + 1 (2 n + 1) !, (6.23) and cosh x = 1 + x 2 2 ! + x 4 4 ! + x 6 6[--=PLGO-SEPARATOR

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  Calculus

  Early Transcendentals

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  3.2 Derivatives of Logarithmic Functions 137 and ln x are the same for x > 0, but ln|x| is defined for all nonzero values of x, and ln x is only defined for positive values of x. Figure 3.2.2 The derivative of ln|x| for x = 0 can be obtained by considering the cases x > 0 and x < 0 separately: Case x > 0. In this case |x| = x, so d dx [ln |x|] = d dx [ln x] = 1 x Case x < 0. In this case |x| = −x, so it follows from (4) that d dx [ln|x|] = d dx [ln (−x)] = 1 (−x) · d dx [−x] = 1 x Since the same formula results in both cases, we have shown that d dx [ln|x|] = 1 x if x = 0 (6) Example 4 From (6) and the chain rule, d dx [ln|sin x|] = 1 sin x · d dx [sin x] = cos x sin x = cot x Logarithmic Differentiation We now consider a technique called Logarithmic Differentiation that is useful for differen- tiating functions that are composed of products, quotients, and powers. Example 5 The derivative of y = x 2 3 √ 7x − 14 (1 + x 2 ) 4 (7) is messy to calculate directly. However, if we first take the natural logarithm of both sides and then use its properties, we can write ln y = 2 ln x + 1 3 ln(7x − 14) − 4ln(1 + x 2 ) Differentiating both sides with respect to x yields 1 y dy dx = 2 x + 7 / 3 7x − 14 − 8x 1 + x 2 Thus, on solving for dy /dx and using (7) we obtain dy dx = x 2 3 √ 7x − 14 (1 + x 2 ) 4  2 x + 1 3x − 6 − 8x 1 + x 2  138 Chapter 3 / Topics in Differentiation REMARK Since ln y is only defined for y > 0, the computations in Example 5 are only valid for x > 2 (verify). However, using the fact that the expression for the derivative of ln y is the same as that for ln |y|, it can be shown that the formula obtained for dy / dx is valid for x < 2 as well as x > 2 (Exercise 59). In general, whenever a derivative dy / dx is obtained by Logarithmic Differentiation, the resulting derivative formula will be valid for all values of x for which y = 0. It may be valid at those points as well, but it is not guaranteed.

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  Anton's Calculus

  Early Transcendentals

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2018(Publication Date)
  • Wiley

    (Publisher)

  3.2 Derivatives of Logarithmic Functions 137 and ln x are the same for x > 0, but ln|x| is defined for all nonzero values of x, and ln x is only defined for positive values of x. Figure 3.2.2 1 −1 x y y = ln⏐x⏐ The derivative of ln|x| for x ≠ 0 can be obtained by considering the cases x > 0 and x < 0 separately: Case x > 0. In this case |x| = x, so d dx [ln |x|] = d dx [ln x] = 1 x Case x < 0. In this case |x| = −x, so it follows from (4) that d dx [ln|x|] = d dx [ln (−x)] = 1 (−x) ⋅ d dx [−x] = 1 x Since the same formula results in both cases, we have shown that d dx [ln|x|] = 1 x if x ≠ 0 (6) Example 4 From (6) and the chain rule, d dx [ln|sin x|] = 1 sin x ⋅ d dx [sin x] = cos x sin x = cot x Logarithmic Differentiation We now consider a technique called Logarithmic Differentiation that is useful for differen- tiating functions that are composed of products, quotients, and powers. Example 5 The derivative of y = x 2 3 √ 7x − 14 (1 + x 2 ) 4 (7) is messy to calculate directly. However, if we first take the natural logarithm of both sides and then use its properties, we can write ln y = 2 ln x + 1 3 ln (7x − 14) − 4 ln (1 + x 2 ) Differentiating both sides with respect to x yields 1 y dy dx = 2 x + 7 ∕ 3 7x − 14 − 8x 1 + x 2 Thus, on solving for dy ∕dx and using (7) we obtain dy dx = x 2 3 √ 7x − 14 (1 + x 2 ) 4 [ 2 x + 1 3x − 6 − 8x 1 + x 2 ] 138 Chapter 3 / Topics in Differentiation REMARK Since ln y is only defined for y > 0, the computations in Example 5 are only valid for x > 2 (verify). However, using the fact that the expression for the derivative of ln y is the same as that for ln |y|, it can be shown that the formula obtained for dy∕dx is valid for x < 2 as well as x > 2 (Exercise 59). In general, whenever a derivative dy ∕dx is obtained by Logarithmic Differentiation, the resulting derivative formula will be valid for all values of x for which y ≠ 0. It may be valid at those points as well, but it is not guaranteed.

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  eBook - PDF

  Calculus Early Transcendentals

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2022(Publication Date)
  • Wiley

    (Publisher)

  1 –1 x y y = ln|x| ▴ Figure 3.2.2 The derivative of ln|x| for x ≠ 0 can be obtained by considering the cases x > 0 and x < 0 separately: Case x > 0. In this case |x| = x, so d dx [ln|x|] = d dx [ln x] = 1 x Case x < 0. In this case |x| = −x, so it follows from (4) that d dx [ln|x|] = d dx [ln(−x)] = 1 (−x) ⋅ d dx [−x] = 1 x Since the same formula results in both cases, we have shown that d dx [ln|x|] = 1 x if x ≠ 0 (6) ▶ Example 4 From (6) and the chain rule, d dx [ln | tan x|] = 1 tan x ⋅ d dx [tan x] = cos x sin x ⋅ [sec 2 x] = 2 cosec 2x Logarithmic Differentiation We now consider a technique called Logarithmic Differentiation that is useful for differen- tiating functions that are composed of products, quotients, and powers. ▶ Example 5 The derivative of y = x 2 3 √ 7x − 14 (1 + x 2 ) 4 (7) is messy to calculate directly. However, if we first take the natural logarithm of both sides and then use its properties, we can write ln y = 2 ln x + 1 3 ln(7x − 14) − 4 ln(1 + x 2 ) Differentiating both sides with respect to x yields 1 y dy dx = 2 x + 7∕3 7x − 14 − 8x 1 + x 2 Thus, on solving for dy∕dx and using (7) we obtain dy dx = x 2 3 √ 7x − 14 (1 + x 2 ) 4 [ 2 x + 1 3x − 6 − 8x 1 + x 2 ] 152 Chapter 3 / Differentiation Remark Since ln y is defined only for y > 0, the computations in Example 5 are valid only for x > 2 (verify). However, using derivative rules shows that the formula obtained for dy∕dx is valid for x < 2 as well as x > 2. In general, whenever a derivative dy∕dx is obtained by Logarithmic Differentiation, the resulting derivative formula will be valid for all values of x for which y ≠ 0. It may be valid at those points as well, but it is not guaranteed (Exercises 61–63). ■ Derivatives of Real Powers of x We know from Theorem 2.3.2 and Exercise 84 in Section 2.3, that the differentiation formula d dx [x r ] = rx r−1 (8) holds for constant integer values of r.

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  Mathematics N5 Student's Book

  TVET FIRST

  • JV John(Author)
  • 2022(Publication Date)
  • Macmillan

    (Publisher)

  Use Logarithmic Differentiation to find dy _ dx : 1.1 y = e x _ ln x (4) 1.2 y = x 1 _ x (4) 1.3 y = (2x + 3) 2x+3 (4) 1.4 y = x tan x (4) 1.5 y = ( 2x − e 8x ) sin 2x [Apr 2019] (4) 1.6 y = x (√ _ x −1) (4) 1.7 y = x ln x _ x 2 (5) 1.8 y = √ _________________________ ( x 2 + 1 )( x 2 + 2 )( x 2 + 3 )( x 2 + 4 ) (5) 2. If y x = x y , use logarithmic and implicit differentiation to prove that dy _ dx = y ( y − x ln y ) _ x ( x − y ln x ) . (6) 3. Use the Logarithmic Differentiation method to verify the product rule: If y = uv, then dy _ dx = v du _ dx + u dv _ dx where u and v are functions of x. (3) 4. Use the Logarithmic Differentiation method to verify the quotient rule: If y = u _ v , then dy _ dx = v du _ dx − u dv _ dx _ v 2 where u and v are functions of x. (3) 53 Differentiation TVET FIRST Module 2 5. If y = (x + 1)(x + 2)(x − 3), show that dy _ dx = 3 x 2 − 7 by differentiating: 5.1 After multiplying out. (4) 5.2 Using the product rule. (4) 5.3 Using the logarithmic method. (4) 6. Determine dy _ dx if y = e x x 2 using Logarithmic Differentiation. [Nov 2012] (4) 7. Determine dy _ dx using Logarithmic Differentiation: 7.1 y = ( tan −1 x ) x (4) 7.2 y = [tan ( tan −1 x )] x [Hint: tan ( tan −1 x ) = x] (4) 7.3 y = (cot x) √ _ x (4) 7.4 y = (e x + x) x (4) 8. Using Logarithmic Differentiation, verify the differentiation rule: If y = a x , then dy _ dx = a x ln a. (2) 9. Determine dy _ dx if x x + y x + 8 x = 16. [Hint: Refer to Example 2.22.] (5) TOTAL: [85] Unit 2.5: Inverse trigonometric functions 2.5.1 Introduction to the inverse trigonometric functions You will remember from N4 that the inverse of a given function is obtained by interchanging the x and the y in its equation. Definition of an inverse function An inverse function is a function that ‘reverses’ another function (undoes the action of another function). f (x) = y ⇔ f −1 (y) = x For example, consider the function y = 2x + 3.

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