Find the derivative of the function.
f(t) = t^3/2 log9(sqrt(t + 5))

the 9 is a small srcipt after the log.

Question

Find the derivative of the function. 
f(t) = t^3/2 log9(sqrt(t + 5))

the 9 is a small srcipt after the log.

zarkon · Answer

this 

$$f(t)=\frac{t^3}{2\log_9(\sqrt{t+5})}$$

anonymous · Answer

no its t^2/3

anonymous · Answer

3/2*

anonymous · Answer

sorry its t^3/2 all over the log stuff

zarkon · Answer

$$f(t)=t^{3/2}\log_9(\sqrt{t+5})$$

anonymous · Answer

yeh

zarkon · Answer

$$=f(t)=t^{3/2}\frac{1}{2}\log_9(t+5)$$

then $$f'(t)=\frac{3}{4}t^{1/2}\log_9(t+5)+\frac{1}{2}t^{3/2}\frac{1}{t+5}\frac{1}{\ln(9)}$$

anonymous · Answer

$$\frac{t^{3/2}}{\log_{9}\sqrt{t+5}} $$?

anonymous · Answer

what he got looks good

anonymous · Answer

no its t^(2/3)

anonymous · Answer

so change the t^3/2 to a 2/3 and do the product rule

zarkon · Answer

$$f(t)=t^{2/3}\frac{1}{2}\log_9(t+5)$$

:)

anonymous · Answer

why si the derivtive so long..somehtign looks wrong:S

anonymous · Answer

1/2 log .. you sure that means logbase9(t+5)/2

anonymous · Answer

whatever zarkon posted thats exactlly how my problem looks

anonymous · Answer

why don't you take a picture of it

anonymous · Answer

zarkon?..whatever zarkon posted is idetical to my problem

anonymous · Answer

so it has a 1/2 in front/

zarkon · Answer

ok...then just modify my first answer

anonymous · Answer

all he's simply doing is the product rule

anonymous · Answer

these are the rules

anonymous · Answer

attachment

anonymous · Answer

ll here is the rule for the last part the first part is simple and you are just going to use the product rule whichis d/dx[uv] = uv'+vu'

anonymous · Answer

i'm not sure how he got u' on the bottom tho

zarkon · Answer

where?

anonymous · Answer

above for the derivative of logbase9(t+5)

anonymous · Answer

o wait u' = 1 nvm

anonymous · Answer

zarkon can u please give me a finla result im so confused...i only have 7 mins left:'(

anonymous · Answer

the answer is above it'd be 2/3 * 1/6

anonymous · Answer

instead

anonymous · Answer

and instead of t^1/2 it'll be t^-1/2

anonymous · Answer

can u write out a final asnwer for me please..my comp keeps freezing

anonymous · Answer

2/3*1/2 which would be 1/3 i meant above

zarkon · Answer

$$t^{-1/3}$$

anonymous · Answer

yes i'm just fumbling aroudn becaause i'm using the numbers from the other equations >_<

anonymous · Answer

zarkonn can u please give me a finla answer..like type it out..my chat screen looks all weird:(

zarkon · Answer

$$f(t)=t^{2/3}\frac{1}{2}\log_9(t+5)$$

$$f'(t)=\frac{1}{2}\left[\frac{2}{3}t^{-1/3}\log_9(t+5)+t^{2/3}\frac{1}{t+5}\frac{1}{\ln(9)}\right]$$

anonymous · Answer

what he has is right

zarkon · Answer

I know ;)

anonymous · Answer

hahaha but you could simplify the back

zarkon · Answer

na...I always tell my students to never simplify.

zarkon · Answer

:)

anonymous · Answer

you're a teacher

anonymous · Answer

?

zarkon · Answer

professor

anonymous · Answer

at?

zarkon · Answer

some college/university in the USA :)