## Derivatives Of Exponential And Logarithmic Functions

Earlier it was claimed that the function e' is its own derivative. As it turns out, the natural log function, In possesses a rather convenient derivative also, namely, d(In t)/dt = \/t. This fact reinforces our preference for the base e. Let us now prove the validity of these two derivative formulas, and then we shall deduce the derivative formulas for certain variants of the exponential and log expressions e' and In t.

Log-Function Rule

The derivative of the log function y = In t is d , 1

To prove this, we recall that, by definition, the derivative of y = f(t) = In / has the following value at t = N:

[by Rule II of logarithms]

N . 1 Now let us introduce a shorthand symbol m =-— .Then we can write t-N--------------t-N

—, and also — = H---— = 1 H--. Thus the expression to the right of the

N N N m limit sign above can be converted to the form

[by Rule III of logarithms]

Note that, when t tends to N, m will tend to infinity. Thus, to find the desired derivative value, we may take the limit of the last expression above as m —> oo:

Since N can be any number for which a logarithm is defined, however, we can generalize this result, and write f'(t) = d(\nt)/dt = \/t. This proves the log-function rule.

Exponential-Function Rule

The derivative of the function^ = e' is

This result follows easily from the log-function rule. We know that the inverse function of the function y = e' is t = In y, with derivative dt/dy = 1 /y. Thus, by the inverse-function rule, we may write immediately dt6 dt dt/dy \/y 'V

### The Rules Generalized

The above two rules can be generalized to cases where the variable t in the expression e' and In t is replaced by some function of t, say, f(t). The generalized versions of the two rules are

The proofs for (10.20) involve nothing more than the straightforward application of the chain rule. Given a functiony = efi'\ we can first let u — f(t), so that y = e". Then, by the chain rule, the derivative emerges as

Similarly, given a function y = In /(/), we can first let v = f(t), so as to form a chain: y = In v, where v = fit). Then, by the chain rule, we have d, n \ d d , dv 1 dv 1 ,

— In fit) = -3-lnc = — lnu — = -— = —~rf (t) dt ' dt dv dt v dt fit) '

Note that the only real modification introduced in (10.20) beyond the simpler rules de'/dt = e' and ¿/(In t )/dt = \/t is the multiplicative factor fit).

Example 1 Find the derivative of the function y = ert. Here, the exponent is rt = fit), with fit) = r\ thus dy _ d dt dt1

Example 2 Find dy/dt from the function y = e '. In this case, fit) = — t, so that fit) = - 1. As a result,

Example 3 Find dy/dt from the function^ = In at. Since in this case fit) = at, with fit) = a, the derivative is d, a 1

— In at = — = — dt at t which is, interestingly enough, identical with the derivative of y = In t.

This example illustrates the fact that a multiplicative constant for t within a log expression drops out in the process of derivation. But note that, for a constant k, we have d , , d k

— A: In t = A: —In / = — dt dt t thus a multiplicative constant without the log expression is still retained in derivation.

Example 4 Find the derivative of the function y = In tc. With fit) = tc and fit) = ctl"\ the formula in (10.20) yields d. c ct£ c

Example 5 Find dy/dt from y = /3ln t2. Because this function is a product of two terms t3 and In t2, the product rule should be used:

dv 1 d , 7 , , d , -f- = t3~In t2 + In t2 — t3 dt dt dt

= It2 + 3i2(21n t) [Rule III of logarithms] = 2;2(1 + 3 In 0

The Case of Base b

For exponential and log functions with base b, the derivatives are

Note that in the special case of base e (when b = e), we have In b = lne = 1, so that these two derivatives will reduce to (d/dt)e' = e' and (d/dt )ln t = \/t, respectively.

The proofs for (10.21) are not difficult. For the case of ft', the proof is based on the identity b = elnh, which enables us to write y _ e(ln b)t _ ^ l In b

(We write rln ft, instead of In ft r, in order to emphasize that t is not a part of the log expression.) Hence

To prove the second part of (10.21), on the other hand, we rely on the basic log property that logA/= (logfc<?)(loge0 = ¿Inr which leads us to the derivative d, d I 1 , \ Id, 1/1

dt &h dt\\nb J In ft dt In b\ t The more general versions of these two formulas are

Again, it is seen that if ft = e, then In ft = 1, and these formulas will reduce to (10.20).

Example 6 Find the derivative of the function/ =12' '. Here, ft = 12,/(;)= 1 - t. and f'(t)= - 1; thus

Higher Derivatives

Higher derivatives of exponential and log functions, like those of other types of functions, are merely the results of repeated differentiation.

Example 7 Find the second derivative of y = b' (with b > 1). The first derivative, by (10.21), is y'(t) = b'\xi b (where In b is, of course, a constant); thus, by differentiating once more with respect to t, we have y"(t) = Jty'{t) = (jtb')[n b = (ft'ln ¿>)ln h = b'(ln bf

Note that y = b' is always positive and In b (for b > 1) is also positive [by (10.16)]; thus y'(t) = b'In b must be positive. And y"(t), being a product of b' and a squared number, is also positive. These facts confirm our previous statement that the exponential function y = b' increases monotonically at an increasing rate.

Example 8 Find the second derivative of y = In t. The first derivative is y' = 1 /t = hence, the second derivative is

Inasmuch as the domain of this function consists of the open interval (0, oo), y' = 1 /t must be a positive number. On the other hand, y" is always negative. Together, these conclusions serve to confirm our earlier allegation that the log function y = In t increases monotonically at a decreasing rate.

### An Application

One of the prime virtues of the logarithm is its ability to convert a multiplication into an addition, and a division into a subtraction. This property can be exploited when we are differentiating a complicated product or quotient of any type of functions (not necessarily exponential or logarithmic).

Example 9 Find dy/dx from x2

Instead of applying the product and quotient rules, we may first take the natural log of both sides of the equation to reduce the function to the form

According to (10.20), the derivative of the left side with respect to x is

4- (left side) = I f dx y dx whereas the right side gives d , . , ., . 2x 1 2 Ix + 6

When the two results are equated and both sides are multiplied by y, we get the desired derivative as follows:

Example 10 Find dy/dx from y = x"ekx ' . Taking the natural log of both sides, we have

In y = a In x + \nekx~c = a\n x + kx — c Differentiating both sides with respect to jc, and using (10.20), we then get and y dx dy

0 0