Differentiation of Natural Logarithm

Suggested Prerequesties:

Properties of exponentials and logarithms, Differentiation of exponentials, Implicit differentiation

The derivative of the natural logarithm (logarithm base e) is one of the most useful derivatives in integral calculus. Even ignoring that, we'd still like to know what it is, in our never-ending quest for knowledge.

We'll go about finding the derivative of the function y = lnx by defining it implicitly, differentiating, and then searching for an explicit solution:

y = lnx

e^y = x

D_xe^y = D_xx

$e^y^dy dx &sp;=&sp; 1$

$dy dx &sp;=&sp; 1 e^y^$

$dy dx &sp;=&sp; 1 x$

So, D_xlnx = x^-1. Polynomials already produce derivatives that are powers of x:

D_x x²/2	= x¹
D_x x¹	= x⁰
D_x (???)	= x^-1
D_x -x^-1	= x^-2
D_x -x^-2/2	= x^-3

and so on. Now we can fill in the ??? with lnx. Probably not what our first guess would have been, but nonetheless true.

You'll notice that I did not explicitly apply the definition of the derivative in order to find D_xlnx. Instead I used other things I already knew about differentiation. Producing this result using the definition would have been much more tedious, and this method is just as correct.

What about logarithms with other bases? Well, since

log_a_x &sp;=&sp; ln x ln a

(why?),

D_x_log_a_x
&sp;=&sp; 1 (ln a)(x)

Then, $D_x_log_e_x &sp;=&sp; 1 (ln e)(x) &sp;=&sp; 1 x$ . That's reassuring, hopefully.

For exponents of bases other than e: a^x = e^(xlna). (why?). Then $D_x_a^x^ &sp;=&sp; D_x_e^x ln a^ &sp;=&sp;
(ln a) e^x ln a^ &sp;=&sp; (ln a)a^x^$

To check again: $D_x_e^x^ &sp;=&sp; D_x_e^x ln e^
&sp;=&sp; (ln e)e^x ln e^ &sp;=&sp; e^x^$

Good.

Some examples:

$D_x_ln (3x) &sp;=&sp; (1 3x) (3) &sp;=&sp; 1 x$
This is an interesting result. It means that the derivative of the log of a constant times a function is equal to the derivative of the function. This may make more sense if we think if it like this:
$D_x_ln (3x) &sp;=&sp; D_x_(ln x + ln 3) &sp;=&sp; 1 x + 0$
$D_x_ln (3x^4^ + 4x +
1) &sp;=&sp; 12x^3^ + 4 3x^4^
+ 4x + 1$
Remember: the chain rule is your friend!
$D_x_log_2_8^x^ &sp;=&sp; (ln 8)8^x^(ln 2)(8^x^) &sp;=&sp; 3 ln 2 ln 2 &sp;=&sp; 3$
To check this: $D_x_log_2_8^x^ &sp;=&sp;
D_x_(log_2_(2^3^)^x^) &sp;=&sp;
D_x_(x log_2_2^3^) &sp;=&sp; D_x_(3x) &sp;=&sp; 3$

Exercises:

Find the derivative with respect to x of each of the following functions:

y = e^xlnx
$h(x) &sp;=&sp; ln (e^x^ + 1) ln (e^x^ - 1)$
g(x) = log₁₀ (ln x)
k(x) = ln (log₁₀ x)
l(x) = (cos2x)^x

Solutions to the exercises | Back to the Calculus page | Back to the World Web Math top page

jjnichol@mit.edu