calculus question?

Question

f(x)= 6^x log_2 (x)

f'(x)=?

Answer 1

f(x) = 6^x log_2(x)

f'(x) = (ln 6) (6^x)(log_2(x)) + (x ln 2)^-1 (6^x)

Product rule

y = f(x)g(x)

y' = f'(x)g(x) + g'(x)f(x)

y = a^x

y' = (ln a) (a^x), a is a constant

log b (x) = y

y' = 1/(ln b x), b is a constant.

Answer 2

First take ln() of both sides

ln[f(x)]= ln[6^x log_2 (x)] = x ln(6) + ln[log_2 (x)]

You can write log_2 (x) = ln(x)/[ln (2)]
Then,
ln[f(x)]= x ln(6) + ln[ln(x)/ln(2)] = x ln(6) + ln[ln(x)] - ln[ln(2)]

Now diff. both sides w.r.t x

[1/f(x)]*f'(x) = ln(6) + [1/ln(x)]*[1/x]

f'(x) = {ln(6) + [1/ln(x)]*[1/x]}f(x)
= {ln(6) + [1/ln(x)]*[1/x]}*[6^x log_2 (x)]

Answer 3

log_2(x) is multiplied or is in the power ?