what is the formla for taking derivative of log(base b) (a)?

Question

i meant "formula"

Answer 1

need to express to base e that is as a natural log

log_b(a) = ln(a) / ln(b)
then
d /da [ ln(a) / ln(b) ] = 1 / [a*ln(b) ]

If a is a function of x then use d /dx and the chain rule

Note be carefull with some earlier answers.

Answer 2

I'm not sure which derivative you want (with respect to a or b?) so we'll derive both.


Using the change of base formula we have

log(base b) (a) = ln(a) / ln(b)


Thus the derivative with respect to a is
1 / [ ( ln(b) ) ( a ) ]


The derivative with respect to b is
- ln(a) / [ b ( ln(b) )^2 ]

Happy Thanksgiving!!!

Answer 3

the derivative...

= [M x (derivative of a)] divided by log(base b) (a)

where M = log(base b) e

e is the base of the natural logarithm(ln)

Answer 4

the derivative of log(base b) of x is ln(b) / x

Answer 5

log(b,a)
log(10,a)/log(10,b)
Derive:
(a'/a)/(b'/b)
a'b/ab'
Where a and b are terms of the derivative. If constant, just cross it out.

Answer 6

It is a constant, its derivative is zero.

Answer 7

d/dxa(log_b(a)= 1/alnb a>0

(in the solution above d/dx log_10( x) is not 1/x, but if you use
ln then the two are the same)