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If you say so.

Thank you!

You're welcome.

\[w=\frac{ \sqrt(x) }{ 2^2+t3^t } \]

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

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 \[w=\frac{ Sqrt(t) }{ 2t^2+t3^t }\]

no you dont use log. i have not learnt that yet

mm maybe but is it possible to go simpler than this?

is it not possible?

no its not possible, other results are

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

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?

right now, I am very confused. ;/ sorry

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)?

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?