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\[w=\frac{ \sqrt(x) }{ 2^2+t3^t } \]
sorry it should be sart (t)
and find dw/dt
that is simple
do u want the answer or do u want a step by step solution?
when u answer my question i sill proceed
Math, you can't simply give them the answer. It's against CoC. Guide them to it, so they can answer it themselves.
step by step pleae!
*please
Quotient rule with respect to (t)..
g(x)f(x)'-f(x)g(x)'/g(x)^2
This right\[w=\frac{\sqrt{x}}{3^t t+2^2}\]
You can use chain rule as well
therefore, (2t62+t3^t)d/dx(sqrt(t)-sqrt(t)(2t^2+t3^t)/(2t^2+t3^t)^2
Followed by product rule
I dont how how to caculate the root.
and how to do (2t62+3t^3) to derivative?
The root "x" is just acted like constant, ignore it
my teacher told me to do something like 1/2x^-1/2? is it wrong
is the question this?\[\Large w=\frac{\sqrt{x}}{3^t t+2^2}\]
no \[w=\frac{ Sqrt(t) }{ 2t^2+t3^t }\]
You sure? the answer is \[\Large \frac{dw}{dt}=-\frac{3^t+t \left(3^t (9 \log )+6\right)}{2 t^{3/2} \left(2 t+3^t\right)^2}\]
no you dont use log. i have not learnt that yet
\[\Large \frac{dw}{dt}=-\frac{3^t+t \left(3^t ( \ln(9) )+6\right)}{2 t^{3/2} \left(2 t+3^t\right)^2}\] This?
mm maybe but is it possible to go simpler than this?
is it not possible?
no its not possible, other results are
1 Attachment
I only have learned power rule, exponential rule and product rule. ( chain rule)
step one use the quotient rule..
the only trick u have here is 3^t, u may use the table or D-operator...
yes I did! its above. I dont know how to break down and power rules.
please show me step by step.
let me show u how to solve 3^t
is that the part ur are struggling with right
i will show u step by step but let's break this down
Yes :( t3^t
thank you so much. I want to cry right now
stop crying and fight for the answer
ok, i wil! thanks
trust me i will let u solve this problem
you wont help me?
so far i have got
damn i wrote the answer on another post hehehe
\[\[u = 3^t, \space du = 3^t \ln(3)\]\]
\[\frac{ (2t^2+t3^t)(\sqrt(x)'-\sqrt(x)(2t^2+t3^t) }{ (2t^2+t3^t)^2 }\]
\[u = 3^t, \space du = 3^t \ln(3)\]
now do u know how this happend
but its t3^t?
forget the t for now because i want to show u bit by bit
because with the t we will get into the product rule
ok, I got it. i am taking note hold on
du_3^t why its In(3)?
good question, now u r starting to listen and that is a good sign of learning...
u= 3^t what happens if you take the natural log of both sides?
mm im not sure, , negative numbers?
ln(u) = ln(3^t)
now do u know how to differentiate both sides?
No I am not sure.
mutiply?
ok no problem, first of all we need some log rules
ln(u) = tln(3)
did u see what happend to t?
power of t went outside of 3 and In?
yes this is a primitive ln rule
Oh I see.
Do i use this everytime i do product?
now if y=ln(x) what is y'?
u mean every time u have a log of a power?
ok, I got it.
sorry power of a log
now if y=ln(x) what is y'?
power of a log, ok !
mmm.. y=In/(x)?
no y=ln(x) ->y'=1/x
wow, thats new to me!
this is basic differentiation of natural log
so now back to our question ln(u) = tln(3) differentiate both sides...
tIn(3)/x
ok let me make this even simple
so u don't get confused
thanks for taking your time.
\[\frac{d}{dt}a^t=a^tln(a)\]
can u memorize this for now and take it for granted
Ok,
ok use this formula and tell me what is 3^t?
hint: a=3
mmm.... 3^tIn(3)?
well done so memorize that please take a note and keep it in ur mind the same way u know ur name...
Ok I will.
whats the next step?
sorry i was replying to another post on set theory ok
its okay.
now tell me how would u differentiate? \[ 2^2+t3^t \]
you mean 2t^2?
2^tIn(2)+3tIn(3)?
no 2^2 is a constant
look at it again
trust me by solving this part we are more than half way of solving the problem
\[2^2+t3^t\]
2^2=4 so when u differentiate a constant u get zero right!?
Oh! so its (4t)?
ok listen let me solve it
can u see how this happened?
using the product rule
No, wheres 2 go? theres no 2 in the answer
\[u=2^2+t3^t \rightarrow du=(3^t+t3^tln3))dt\]
listen y=4, what is y'?
y=-4?
zero? beacuse its constant
zero well done, so y=2^2, y'=?
zero?
yes well done
so differentiate now t3^t
3^3In(t)?
no its 3 ^tIn(3)?
\[u=2^2+t3^t \rightarrow du=(3^t+t3^tln3))dt\]
do u know the product rule?
I am not sure
uv'+u'v???
do u know that rule?
yes I do qutiotinent rule?
no i said product rule..
do u want me to solve the whole thing then u may go and think about it?
sorry no.
Yes please,
the whole thing and if i have a question can i ask?
then how come u are solving such a problem when u need the fundamentals?
I have rules here but im confused how i am apply to . I am able to solve the basic problems
i have test next week and I want to do good :(
please help me with this question.
I am and i will i promised
Thank you so much.
can u memorize the followings for me ...!
if u have u x v then when u want to differentiate it we use something called the product rule...
Ok thank you for the fact.
example :
if u= x^2 and v = e^x, and we were asked to find the derivative of(uv)
we apply this rule u v' + u' v
ok!
so what u do is u=x^2, u'=2x, v=e^x, u=e^x
now can u put the above in the following u v' + u' v
mm for my question?
no for my question
if u= x^2 and v = e^x, and we were asked to find the derivative of(uv) so what u do is u=x^2, u'=2x, v=e^x, u=e^x u v' + u' v
ok, x^2(e^x)+2x(ve^x)
well done but without the v
without v in the end?
look at what u wrote
yes,
what is the v doing in the equation and u got rid of it by subbing the functions in terms of x
Oh! sorry yes.
ok good now solve for ur question t3^t
how?
using the product rule then one u just solved above
so use the same steps i showed u
\[f(x)(2t^2+t3^t)(sqrtx)-(sqrtx)(2t^2+t3^t)\] from the producet rule?
no i said solve this one t2^t, please follow what i am asking u to do...not ur own rules or derrieres
but thats what i have told in the class something like that. or maybe I was wrong. could you show me everything and the answer?
right now, I am very confused. ;/ sorry
yes but don't forget we decided to break it a part then when we join everything together u'll see the final answer
O. I see sorry i was misleading
ok then lead now, so wat is diff of t3^t?
(t)(3^t)"+t?
use the u v' + u' v rule that i just showed u
(t)(3^t)+(3^t)(t)?
i just told u how to find u' of 3^t
i told u to memorize or take a not of it
\Large w=\frac{\sqrt{t}}{3^t t+2^2}
\[\Large w=\frac{\sqrt{t}}{3^t t+2^2}\]
this is ur equation right?
no.
its 2t^2+t3^t
oh ok
\[\Large w=\frac{\sqrt{t}}{3^t t+2t^2}\]
ok let's us apply the quotient rule
so let u =t^(1/2) and v=2t^2+t3^t, so what would u', and v' be?
or what would du=? and dv=?
uv'=vu"
(t^1/2)(2t^2+t3^t)+(2t^2+t3^t)
u = t^(1/2) that mean u'=(1/2)t^(-(1/2))
yes, i got it
ops sorry
good girl now tell me
if v=2t^2+t3^t what is v'?
how did you guess that i was a girl. 3^2in(3)?
why do u think i am wrong?
?
\[v=2t^2+t3^t \rightarrow v'=4t+3^t+t(3^tln(3))\]
are u there or surprise?
sorry i was looking above to recall again.
i told u to write
write on a piece of paper
I am.
\[\frac{g(x)f(x)'-f(x)g(x)'}{g(x)^2}\]
yes,
\[g(x)=\sqrt{t} \space and \space f(x)=2t^2+t3^t\]
plug into the equation right?
yes
but f'(x) and g"(x) needs to be calculated differently
yes
g(x) =t^(1/2), what is g'(x)?
1/2t^-(1/2)
how?
nope try again
recall \[nt^{(n-1)}\]
hint n=(1/2)
r u there?
(1/2) t^-(1/2)
well done
if u solve this one the ur almost done with the question
\[ f(t)=2t^2+t3^t \space f'(t)?\]
are u there?
t3^t(in3)?
what happened to the 2t^2 and the product rule, if u look up i solved this for like 23084920384 times
4t+2^t+t3^t(In3)
good now plug the whole thing in the quotient rule
Oh I see thank you! i got it~
:)
ya but that is not the end
u still have more to the problem
sorry could you wait for 5 mins? im going back to my room and change to my comp?

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