find f'(x) of f(x)=logax base a + lnax where a is constant.

Asked on by sylli

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

It is given that f(x) = log(a) x + ln a*x

f(x) = log(a) x + ln a*x

use the property of logarithm: log(a) x = ln x / ln a and ln a*x = ln a + ln x

=> f(x) = ln x/ ln a + ln a + ln x

=> f'(x) = (1/x*ln a) + 1/x

=> f'(x) = (1/x)(1 + 1/ln a)

=> (ln a +1)/((ln a)*x)

The required derivative is f'(x) = (ln a +1)/((ln a)*x)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The derivative of the given function is:

f'(x) = [log a (x) + ln (ax)]'

f'(x) = [log a (x)]' + [ ln (ax)]'

We'll re-write the first term of the expression, changing the "a" base into the natural base:

log a (x) = ln x/ln a

We'll differentiate with respect to x:

ln x/ln a = 1/x*ln a = 1/ln (a^x)

[ ln (ax)]' = (ax)'/ax = a/ax = 1/x

The requested derivative of the function is:f'(x)=1/ln (a^x) + 1/x

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