Wednesday

July 30, 2014

July 30, 2014

Posted by **nicole** on Wednesday, October 31, 2012 at 11:38am.

first one:

f(t)=2^(log5t)

the 5 is the base

second one:

y=(2(x^2) - 1)^5 /(√(x+1)

for the second one you have to use logarithmic differentiation

- calculus -
**Steve**, Wednesday, October 31, 2012 at 12:19pmrecall 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)

- calculus -
**Steve**, Wednesday, October 31, 2012 at 12:20pmoops. forgot an ^4 on the last line

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