Calc 1
Tues 10/7/08
4.2 - Derivatives of Log functions
Well, if we’re going to ▯gure out what the derivative of ln(x) is, we might as well get cracking!
d ln(x + h) ▯ ln(x)
[ln(x)] = lim
dx h!0 h
1
= lim [ln(x + h) ▯ ln(x)]
h!0 h ▯ ▯
1 x + h
= lim ln
h!0 h ▯ x ▯
1 h
= lim ln 1 +
h!0 h x
h
Okay, now let’s do a substitution. Letxv =Notice that if we think of x as ▯xed, then when we
take h ! 0, then v ! 0. This means
1
= lim ln(1 + v)
v!0vx
1 1
= lim ln(1 + v)
xv!0 v
1 1=v
= lim ln(1 + v)
xv!0
Oh, but we’ve seen that limit before! ...Or at least a variation of it. We’ve seen
▯ ▯
1 x
lim 1 +
x!1 x
which is exactly the same thing. What’s that limit? Why, it’s e! So, our limit is just
1 1
lim ln(e) =
x v!0 x
All right, so[ln(x)] =1, so long as x > 0. Also, for any base b,
dx x
d [log (x)] = 1
dx b xln(b)
Example: ▯nd d [ln(2x + 3)].
dx
Solution: This is just chain rule, so derivative of the outside woul, and then multiply
2x+3
that by the derivative of the inside, which is 2. So, the derivative will be
2
2x + 3
Since it’s so much easier taking the derivatives of sums and di▯erences rather than products and
quotients (because you have to use those rules!), a lot of times, it’s much better to use log
properties to split up the functions into sums and then take the derivative.
1 Example:
▯ ▯
d x sin(x) d ▯ ▯
ln 2 = ln(x ) + ln(sin(x)) ▯ ln(tan(x + 1))
dx tan(x + 1) dx
2 2 2
= 3x + cos(x)▯ 2xsec (x + 1)
x3 sin(x) tan(x + 1)
2 2
3 2xsec (x + 1)
= x + cot(x) ▯ tan(x + 1)
Oh, and by the way, you can extend the derivative of log functions to
d 1
lnjxj = ; x 6= 0
dx x
This will become more important when we do integrals later on. But for now, let’s move on to

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