How do you differentiate #log_2 (x)#?
This follows from the general formula:
As we know how to differentiate
The according formula to change a logarithmic expression from the base
#log_color(red)(a)(color(blue)(x)) = log_b(color(blue)(x)) / log_b(color(red)(a)) #
You can apply the formula as follows:
#log_2(x) = ln(x) / ln(2)#
#"d"/("d"x) [log_2(x)] = "d"/("d"x) [ln(x) / ln(2) ] = 1/ln(2) * 1/x = 1/(xln(2))#