Logarithm Functions : Now let's briefly get the derivatives for logarithms. In this case we will have to start with the following fact regarding functions that are inverses of each other.
Fact 2 : If f(x) & g(x) are inverses of each other then,
g′ ( x ) = 1/ f ′ ( g ( x ))
Hence, how is this issue useful to us? Well recall that the natural exponential function and the natural logarithm function are inverses of each other and we know derivative of the natural exponential function.
Hence, if we have f ( x ) = ex and g ( x ) =ln x then,
g′ ( x ) = 1/f ′ ( g ( x )) = 1/ e g ( x ) = 1/ e ln x =1/ x
The final step just utilizes the fact that the two functions are inverses of each other.
Putting this all together gives,
d (ln x )/dx = 1/x x>0
Note as well that we have to require that x > 0 as it is required for the logarithm and hence have to also be needed for its derivative. It can also be illustrated that,
d(ln |x| ) /dx= 1/x x ≠ 0
By using this all we have to avoid is x=0
In this, unlike the exponential function case, actually we can determine the derivative of the general logarithm function. All that we required is the derivative of the natural logarithm, that we only found, and the change of base formula. By using the change of base formula we may write a general logarithm as,
loga x = ln x /ln a
Differentiation is then fairly simple.
d(log a x)/dx = d(lnx/lna)/dx
= (1/lna )(d(lnx)/dx
= 1/xlna
We took benefit of the fact that a was a constant and thus ln a is also a constant and can be factored of the derivative. Putting all of this together gives,
d (logax)/dx =1/(xlna)
Following is a summary of the derivatives in this section.
d (ex )/dx= ex d (a x ) / dx = a x ln a
d (ln x ) /dx= 1 d(log a x)/dx = 1/xlna