Posted by nicole on Wednesday, October 31, 2012 at 11:38am.
recall that (a^u)' = lna a^u u', so we have
f' = ln2 2^log5t (log5t)'
Now, log5t = lnt/ln5, so
f' = ln2 2^log5t * 1/ln5 * 1/t
f' = ln2/ln5 * 1/t 2^log5t
Now, ln2/ln5 = log5(2), so finally,
f' = (log5(2) / t) 2^log5t
y = u^5/v where
u = 2x^2-1
v = √(x+1)
y' = (5u^4 u' v - u^5 v')/v^2
= u^4 (5vu' - v')/v^2
= (2x^2-1)^4 (5(2x^2-1) - 1/(2√(x+1)))/(x+1)
You can massage this as you want; one form is
(2x^2-1) (38x^2+40x+1)
------------------------------
2(x+1)√(x+1)
oops. forgot an ^4 on the last line
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