3.6 Derivatives of Logarithmic Functions Recall how to differentiate inverse functions using implicit differentiation. Since the natural loga- rithm is the inverse function of the natural exponential, we have

dy dy 1 1 y = ln x ey = x = ey = 1 = = = ⇐⇒ ⇒ dx ⇒ dx ey x y We have therefore proved the first part of the following The- 3 orem: the remainder follow immediately using the log laws and chain rule. 2 y = 1 d 1 x Theorem. If x > 0 then ln x = • dx x 1 d 1 If x = 0, then ln x = • 6 dx | | x 3 2 1123 − − − d d ln x 1 1 x log x = = − y = ln x • dx a dx ln a x ln a | | 2 d d − ax = ex ln a = (ln a)ex ln a = (ln a)ax • dx dx 3 − The last two parts of the Theorem illustrate why calculus always uses the natural logarithm and expo- nential. Any other base causes an extra factor of ln a to appear in the derivative. Recall that ln e = 1, so that this factor never appears for the natural functions.

Example We can combine these rules with the chain rule. For example:

d 1 d 2x log (x2 + 7) = (x2 + 7) = dx 4 (x2 + 7)(ln 4) · dx (x2 + 7)(ln 4)

Logarithmic Differentiation This is a powerful technique, allowing us to use the log laws to simplify an expression before differ- entiating. For example, suppose that you wanted to differentiate

3x2 + 1 g(x) = ln √1 + x2 Blindly applying the chain rule, we would treat g as ln(lump), then be stuck with a hideously un- pleasant quotient rule calculation:

d 3x2 + 1 1 d 3x2 + 1 √1 + x2 d 3x2 + 1 g0(x) = ln = 2 = dx √1 + x2 3x +1 dx √1 + x2 3x2 + 1 dx √1 + x2 √1+x2 √1 + x2 6x√1 + x2 (3x2 + 1) x(1 + x2) 1/2 = − · − 3x2 + 1 · 1 + x2 1 6x(1 + x2) (3x2 + 1) x x(3x2 + 5) = − · = 3x2 + 1 · 1 + x2 (3x2 + 1)(1 + x2)

1 Thankfully there is a better way: apply the logarithm laws first,

3x2 + 1 1 g(x) = ln = ln(3x2 + 1) ln(1 + x2) √1 + x2 − 2 and then differentiate:

1 d 2 1 d 2 g0(x) = (3x + 1) (1 + x ) 3x2 + 1 dx − 2(1 + x2) dx 6x x = 3x2 + 1 − 1 + x2 A little algebra shows that we have the same solution, in a much simpler way.

Logarithmic differentiation is so useful, that it is most often applied to expressions which do not contain any logarithms at all. Suppose instead that we had wanted to differentiate

3x2 + 1 f (x) = √1 + x2 Then g(x) = ln f (x) is easy to differentiate and, since

f 0(x) g0(x) = = f 0(x) = f (x)g0(x) f (x) ⇒ we can immediately write the derivative:

3x2 + 1  6x x  f 0(x) = √1 + x2 3x2 + 1 − 1 + x2

Logarithmic Differentiation in General To differentiate y = f (x) using logarithmic differentiation. 1. Take natural logs of both sides: ln y = ln f (x)

1 dy 2. (Implicitly) Differentiate with respect to x: the left hand side becomes y dx

dy 3. Solve for dx . This is especially useful if the form of f (x) is something that simplifies nicely under the logarithm laws: for example products, quotients and powers.

Examples (2x + 1)1/3(1 + x2) 1. Let y = . Then (2 + x4)2/5 1 2 ln y = ln(2x + 1) + ln(1 + x2) ln(2 + x4) 3 − 5 1 dy 2 2x 8x3 = = + ⇒ y dx 3(2x + 1) 1 + x2 − 2 + x4 dy (2x + 1)1/3(1 + x2)  2 2x 8x3  = = + ⇒ dx (2 + x4)2/5 3(2x + 1) 1 + x2 − 2 + x4

2 2. Let y = (sin x)√x. Then ln y = √x ln(sin x), from which

1 dy 1 1/2 cos x = x− ln(sin x) + √x (product rule) y dx 2 · sin x 1 = [ln(sin x) + 2 cot x] 2√x dy y (sin x)√x = = [ln(sin x) + 2 cot x] = [ln(sin x) + 2 cot x] ⇒ dx 2√x 2√x e as a limit
n We have already seen the suggestion that lim 1 + 1 = e when we discussed compound interest. n ∞ n Here is a proof, using the definition of derivative.→ Let f (x) = ln x, then

1 d f ln(x + h) ln x  x + h 1/h = = lim − = lim ln x dx h 0+ h h 0+ x → → 1 1 Now write n = h and y = x to obtain  y n  y n  y n y = lim ln 1 + = ln lim 1 + = ey = lim 1 + n ∞ n ∞ n ∞ → n → n ⇒ → n This is usually written as  x n ex = lim 1 + or lim (1 + x)1/x = e n ∞ n x 0 → → Homework

y x dy 1. If x = y , use implicit and logarithmic differentiation to find dx . 2. Suppose that a is constant and the functions f and g are related by

f (x) = ag(x)

g(x) Prove that f 0(x) = (ln a)g0(x)a . 3. Similarly to the previous question, compute

d (g(x))h(x) dx

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