What is the derivative of #y=log_10(x)#?

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Answer 1 · 2014-07-25T16:56:12

The answer is

#y'=log_10(e)*1/x#

Solution

Suppose we have #log_a(b)#, we want to change it on exponential (e) base, then it can be written as:

#log_a(b)=log_a(e)*log_e(b)#

Similarly, function #log_10(x)# can be written as:

#y=log_10(e)*log_e(x)#

Let's say we have, #y=c*f(x)#, where c is a constant
then, #y'=c*f'(x)#

Now, this is quite straightforward to differentiate, as #log_10(e)# is constant, so only remaining function is #log_e(x)#

Hence:

#y'=log_10(e)*1/x#

Alternate solution:

Another common approach is to use the change of base formula, which says that:

#log_a(b) =ln(b)/ln(a)#

From change of base we have #log_10(x) = log_10(x) = ln(x)/ln(10)#.

This we can differentiate as long as we remember that

#1/ln(10)# is just a constant multipler.

Doing the problem this way gives a result of #y' = 1/ln(10)*1/x#.