Lnx) = Lim Ln(X + H

Home , Logarithmic derivative

Differentiating logarithm functions. To differentiate y = ln x, we must return to the definition... d ln(x + h) − ln x (ln x) = lim dx h→0 h and simplify, using algebraic properties of ln x... 1 = lim (ln(x + h) − ln x) h→0 h 1 x + h = lim ln h→0 h x " # h1/h = lim ln 1 + h→0 x and use the continuity of ln x... " # h1/h = ln lim 1 + h→0 x

What next?

1 Remember the special limit lim (1 + u)1/u = e . . . u→0 and substitute (rename) h 1 1 1 1 = u =⇒ h = ux =⇒ = = · ... x h ux u x So h1/h 1 1 lim 1 + = lim (1 + u) u · x h→0 x u→0 1 1 x = lim (1 + u) u u→0 1 1 x = lim (1 + u) u u→0 (because f(x) = a1/x is continuous) = e1/x

2 d Returning to (ln x)... dx d ln(x + h) − ln x (ln x) = lim dx h→0 h . . " # h1/h = ln lim 1 + h→0 x = ln(e1/x) 1 = . x I.e., d 1 (ln x) = dx x

3 Example 1. d 1 3x2 ln x = 6x ln x + 3x2 · = 6x ln x + 3x dx x (product rule) Example 2. d 1 10x + 3 ln 5x2 + 3x + 1 = · (10x + 3) = dx 5x2 + 3x + 1 5x2 + 3x + 1 (chain rule) Example 3. Diﬀerentiate y = ln 5x2. We can use the chain rule again: 1 10x 2 y0 = · 10x = = ... 5x2 5x2 x or we can simplify and then diﬀerentiate: 1 2 y = ln 5x2 = ln 5 + ln x2 = ln 5 + 2 ln x =⇒ y0 = 0 + 2 · = x x

4 Example 4. Find the equation of the tangent line to the graph x2 + x − 1 y = ln 4x − 3 at the point where x = 1. 1 + 1 − 1 Point: y(1) = ln = ln 1 = 0, so the point is (1, 0). 4 − 3 Slope: We need to compute y0(1), and once again, it is easier to simplify before we diﬀerentiate: x2 + x − 1 y = ln = ln(x2 + x − 1) − ln(4x − 3) 4x − 3 2x + 1 4 =⇒ y0 = − x2 + x − 1 4x − 3 3 4 =⇒ y0(1) = − = −1. 1 1 So the equation of the tangent line is

y = y(1) + y0(1)(x − 1) =⇒ y = −(x − 1) =⇒ y = 1 − x.

5 x2+x−1 Figure 1: Graph of y = ln 4x−3 and its tangent line at (1, 0)

6 Logarithmic differentiation: If f(x) is any (differentiable) function, then d 1 f 0(x) ln(f(x)) = · f 0(x) = . dx f(x) f(x) This is called the logarithmic derivative of f(x). d In some cases, it is easier to compute the dx ln(f(x)) than it is to compute f 0(x), because ln(f(x)) is a simpler function than f(x) itself. We can take advantage of this to find f 0(x) because d f 0(x) = f(x) · ln(f(x)) . dx This is called logarithmic differentiation. We will use this idea in the following special, but important case...

7 Differentiating the exponential function. Observation: ln(ex) = x which is a simpler function (to differentiate) d than ex. So we will use logarithmic differentiation to find ex: dx d d d ex = ex · (ln(ex)) = ex · x = ex · 1 = ex. dx dx dx I.e., d ex = ex dx Example 5. d 1 3 3exx1/2 = 3exx1/2 + 3ex · x−1/2 = 3exx1/2 + exx−1/2 dx 2 2 product rule. Example 6. 2 0 2 ex −3x+1 = ex −3x+1(2x − 3) chain rule

8 Example 7. Find the marginal revenue function for the ﬁrm whose demand equation is given by

p = 15e2−0.05q.

First, ﬁnd the revenue function

r = pq = 15qe2−0.05q.

Now diﬀerentiate, using the product and chain rules: dr = 15e2−0.05q + 15qe2−0.05q(−0.05) = (15 − 0.75q)e2−0.05q. dq

9 The consumption function for a small country is given by e0.95Y C = ln , e0.2Y + 5 where Y is national income, measured in $billions. (a) How much is consumed when Y = 10? e9.5 C(10) = ln ≈ 6.983 e2 + 5 (b) What is the marginal propensity to consume when Y = 10? dC d e0.95Y = ln dY dY e0.2Y + 5 d = ln e0.95Y − ln e0.2Y + 5 dY d 0.2e0.2Y = 0.95Y − ln e0.2Y + 5 = 0.95 − dY e0.2Y + 5 2 dC 0.2e = 0.95 − 2 ≈ 0.8307 dY Y =10 e + 5

10 dC (c) Compute the limit lim , and interpret the result. Y →∞ dY dC 0.2e0.2Y lim = lim 0.95 − Y →∞ dY Y →∞ e0.2Y + 5 0.2e0.2Y · e−0.2Y = 0.95 − lim Y →∞ (e0.2Y + 5) · e−0.2Y 0.2 0.2 = 0.95 − lim = 0.95 − = 0.75 Y →∞ 1 + 5e−0.2Y 1 Interpretation: When income grows large, the nation will tend to consume $0.75 of each additional dollar of income.