Question

\[u^3\sqrt(t)+ 2 \sqrt(t^3)\]

Answer 1

yes what are you trying to do ? :)

Answer 2

derivative sorry

Answer 3

Differentiate the sum term by term, this requires power rule and chain rule,

\[u^3\frac{ d }{ dt }(\sqrt{t})+2\frac{ d }{ dt }\sqrt{t^3}\] I just factored the constants out for now, can you do this?

Answer 4

It might be easier for you if you convert the square root to exponents \[\sqrt{t} = t ^{1/2}\]

Answer 5

But hey, I'm going to sleep now, @ganeshie8 mind taking over :3?

Answer 6

wait u^3 isnt a constant isnt it?

u^3 1/2t^(-1/2) + 2 * 1/2t^(-1/2) * 3t^2

Answer 7

Is it given that \( u\) is a function of \(t\) ?

Answer 8

you're right if \(u\) is a constant, otherwise simply use chain rule for the first product

Answer 9

product rule*
\[\large \left(fg\right)' = f'g + fg'\]

Answer 10

oh, im sorry. i did the question wrong i still need help but instead of u to the 3rd its a different problem let me rewrite it

Answer 11

\[u=\sqrt[3]{t^2}+2\sqrt{t^3}\]

Answer 12

let me rephrase the problem :
\(\large f(u,t) = u^3\sqrt{t}+ 2 \sqrt{t^3}\)

\(\large u=\sqrt[3]{t^2}+2\sqrt{t^3}\)


\(\large \dfrac{df}{dt} = ?\)

Answer 13

is that the problem ? or have i mixed both the problems :o

Answer 14

no you mixed both problems, the only problem is the u= cuberoot problem the last one i posted im sorry if i confused you

Answer 15

haha i thought so ! so you just want to find \(\large \dfrac{du}{dt}\) given this :
\(\large u=\sqrt[3]{t^2}+2\sqrt{t^3}\)

?

Answer 16

yes!

Answer 17

good, we're on same page finally :) so take the derivative, whats stopping you ?

Answer 18

\(\large u=\sqrt[3]{t^2}+2\sqrt{t^3}\)

\(\large \dfrac{d}{dt}(u) = \dfrac{d}{dt} \left(\sqrt[3]{t^2}+2\sqrt{t^3}\right)\)

Answer 19

i dont really understand how to do it. it becomes a mess. how would i go about taking the derivative of (t^2)^1/3 would you use the chain rule?

Answer 20

write \(\large \sqrt[3]{t^2}\) as \(\large t^{\frac{2}{3}}\) and use the power rule :

\(\large \dfrac{d}{dx}(x^n) = nx^{n-1}\)

Answer 21

\(\large u=\sqrt[3]{t^2}+2\sqrt{t^3}\)

\(\large \dfrac{d}{dt}(u) = \dfrac{d}{dt} \left(\sqrt[3]{t^2}+2\sqrt{t^3}\right)\)

\(\large ~~~~~~~~~~~ = \dfrac{d}{dt} \left(t^{\frac{2}{3}}+2t^{\frac{3}{2}}\right)\)

Answer 22

2/3t^(-1/3) + 3/2t^(1/3) ?

Answer 23

looks perfect !

Answer 24

wait a sec, who ate the \(2\) that was attached to the second term before differentiating ?

Answer 25

i forgot the 2 lol it'd be the 2/3^(-1/3) + 2(3/2(^1/3)

Answer 26

yes ! we're done with the derivative part. calculus part is over... you may leave the answer as it is or simplify it - up to you :)

Answer 27

i need help simplfying it thank you! i dont really understand how to simplify negative fractions or fractional exponents

Answer 28

sure, so where are we

Answer 29

\(\large ~~~~~~~~~~~ = \dfrac{d}{dt} \left(t^{\frac{2}{3}}+2t^{\frac{3}{2}}\right)\)

\(\large ~~~~~~~~~~~ = \frac{2}{3}t^{-\frac{1}{3}} + 2\frac{3}{2}t^{\frac{1}{3}}\)

\(\large ~~~~~~~~~~~ = \frac{2}{3}t^{-\frac{1}{3}} + 3t^{\frac{1}{3}}\)

Answer 30

fine so far ? :)

Answer 31

mhm everythings perfect so far

Answer 32

somehow teachers don't like negative exponents, so we use this exponent rule :
\[\large a^{-m} = \dfrac{1}{a^m}\]

Answer 33

\[\large ~~~~~~~~~~~ = \frac{2}{3}t^{-\frac{1}{3}} + 3t^{\frac{1}{3}}\]

\[\large ~~~~~~~~~~~ = \frac{2}{3}\dfrac{1}{t^{\frac{1}{3}}} + 3t^{\frac{1}{3}}\]

Answer 34

\[\large ~~~~~~~~~~~ = \frac{2}{3\sqrt[3]{t}} + 3\sqrt[3]{t}\]

Answer 35

guess i would stop here..

Answer 36

ohh is that really it? thank you very much youve been a huge help!

Answer 37

thats rly it ! yw :)