recall that
y = loga u
y' = (1/lna)*(1/u)*(du/dy)
I will assume that you have
r(t) = log4 (t + √t)
r ' (t) = (1/ln4)*( 1/(t + √t) )*(1 + (1/2)t^(-1/2) )
= (1/ln4) (1 + 1/(2√t) )/(t + √t)
if we multiply top and bottom by 2√t
= (1/ln4) (2√t + 1)/(2t^(3/2) + 2t)
or variations of that
Find the derivative of the function.
r(t) = log_4(t + sqrt(t))
My answer (sorry, it's messy):
(1+(1/2sqrt(t)))(1/tln(4)+sqrt(t)ln(4))
3 answers
As I read your solution, it appears to be correct.
I think this is what you meant, the second term is the denomiator... (1+(1/2sqrt(t)))/( (tln(4)+sqrt(t)ln(4)) )
I think this is what you meant, the second term is the denomiator... (1+(1/2sqrt(t)))/( (tln(4)+sqrt(t)ln(4)) )
Yes, that's what I meant. :) Thank you both!