How do you differentiate #f(x)= sqrt (ln2^x)#?

1 Answer
George C.
Nov 15, 2015

Use properties of #ln# and power rule to find:

#d/(dx) sqrt(ln 2^x) = 1/2 sqrt(ln 2^(1/x))#

Explanation:

Use properties of #ln# to note that #ln(2^x) = x ln(2)#.

Use power rule to differentiate #x^(1/2)#

Reformulate in terms of #2^(1/x)#

So:

#d/(dx) f(x) = d/(dx) sqrt(ln 2^x)#

#= d/(dx) sqrt(x ln(2))#

#=d/(dx) (x ln(2))^(1/2)#

#= sqrt(ln(2)) d/(dx) x^(1/2)#

#= sqrt(ln(2)) * 1/2 x ^(-1/2)#

#= 1/2sqrt(ln(2)/x)#

#= 1/2 sqrt(ln 2^(1/x))#

Related questions
See all questions in Differentiating Logarithmic Functions with Base e
Impact of this question
1287 views around the world
You can reuse this answer
Creative Commons License