Question

https://i.gyazo.com/08bf7f7ace4ae7411ab706e02b2b16bb.png
Starting with question b. I'm confused on how to find the derivative.

Answer 1

@Vocaloid

Answer 2

@dude

Answer 3

@Shadow

Answer 4

@mhchen @Vocaloid @Hero
Those are your active calculus helpers. I am afraid I am not knowledgeable in this area.

Answer 5

Okay, thanks.

Answer 6

\[f(x)= x^2\] can you take the derivative of this function ??
are you familiar with the derivative rules?

Answer 7

Yes.
2x

Answer 8

good

Answer 9

I got this as the derivative but not sure
(1-bx+bxRT)/(1-bx^2) - 2ax^2

Answer 10

so for your question it says RT are constant \[P(x)' =\frac{ d }{ dx }(\frac{ Kx }{ 1-bx })-\frac{ d }{ dx }(ax^2)\]

Answer 11

what is the derivative of ax^2 ?

Answer 12

I wanted to say 2x, but not sure tbh

Answer 13

And why do you have Kx? Shouldn't it be RTx ?

Answer 14

yes to make it less confusing i replaced "RT" with K for constant lel

Answer 15

Oh lol

Answer 16

\(\color{#0cbb34}{\text{Originally Posted by}}\) @AbuZZ
I wanted to say 2x, but not sure tbh
\(\color{#0cbb34}{\text{End of Quote}}\)
what happened to a ??

Answer 17

That's why I originally said -2ax^2

Answer 18

we are taking derivative with respect to x so we have assume all other variables are constant.

Answer 19

But when you re asked the question I got second thoughts lol

Answer 20

\[\frac{d}{dx}(ax^{\color{Red}{2}})= \color{Red}{2}ax^{\color{red}{2}-\color{blue}{1}}=\color{red}{2}ax \]
so you drop the power and subtract one from the exponent

Answer 21

Sorry, yes I meant 2ax my bad

Answer 22

alright now for the first part \[\frac{ RTx }{ 1-bx }\]
apply quotient rule
\[(\frac{ f }{ g})'=\frac{f'g -fg'}{g^2}\]

Answer 23

\(\color{#0cbb34}{\text{Originally Posted by}}\) @AbuZZ
I got this as the derivative but not sure
(1-bx+bxRT)/(1-bx^2) - 2ax^2
\(\color{#0cbb34}{\text{End of Quote}}\)
are you sure you typed it correctly ?

Answer 24

Here lemme do it in steps

The derivative of RT(x) = 1
The derivative of 1-bx = -b
So:
(1(1-bx)-RTx(-b))/(1-bx)^2

Answer 25

Simplifying that we get
(1-bx+bxRT)/(1-bx)^2 -2ax

Answer 26

hmm let me replace "RT" with a number RT= 2
just to make easier
\[\frac{ 2x }{ 1-bx }= 2\color{Red}{ \frac{ d }{ dx}(\frac{ x }{ 1-bx})}\]
now take the derivative
\[\color{red}{\frac{ 1(1-bx)-(x)(-b) }{ (1-bx)^3 }}\]

Answer 27

thats seems right i guess typing error ??
`(1-bx+bxRT) `/(1-bx)^2 -2ax
should be `RT(1-bx+bx)`/(1-bx)^2

Answer 28

Why ^3?

Answer 29

^2 ***

Answer 30

Okay.
So what happened to the 2 for the RT on the top?

Answer 31

it stays there
red highlighted one is just the derivative of the fraction
so when i have to apply the quotient or product rule and if there is constant next to the function i like to take it out

Answer 32

Ok.
But I don't get it, did I do something wrong?

Answer 33

\[\frac{ 2x }{ 1-bx }= 2\color{Red}{ \frac{ d }{ dx}(\frac{ x }{ 1-bx})}\]
\[\color{red}{\frac{ 1(1-bx)-(x)(-b) }{ (1-bx)^3 }}\]
\[\rm \color{red}{\frac{ 1(1-bx)-(x)(-b) }{ (1-bx)^3 }}\]
\[RT\frac{(1\cancel{-bx}+\cancel{bx}) }{(1-bx)^2 } -2ax\]

Answer 34

just be careful with parenthesis
RT(1-bx+bx) isn't the same as (1-bx+bxRT)

Answer 35

Okay so we get
RT/1 x 1/(1-bx)^2 - 2ax?

Answer 36

\[\frac{ RT }{ 1 } \times \frac{1}{(1-bx)^2} -2ax \] yes which can be written as
\[\frac{ RT }{ (1-bx)^2 }-2ax\]

Answer 37

But I don’t get it, how can you factor out RT when RT is only present in one of them? Or are you treating the whole numerator as one?

Answer 38

no i didn't factor out rt from the entire function
\[\rm RT (\frac{ 1 }{ (1-bx)^2 }) = \frac{ RT }{ (1-bx)^2}\]
if it was in both terms then this is how we suppose to write \[\rm RT (\frac{ 1 }{ (1-bx)^2 } -2ax)\]

see the difference ??

Answer 39

Yeah, I see that part but from this stage
(1-bx+bxRT)/(1-bx)^2

Answer 40

parenthesis plays a major rule here
if you know where to put them and when it will leads you to the correct answer

Answer 41

Did I do it correctly? :\

Answer 42

can you post step by step solution
because i thought that was an error typing bxRT together

Answer 43

Okay so the original function:
p(x) = (RT(x))/(1-bx)-ax^2
P'(x) = (1(1-bx)-(-b)(RT(x)))/(1-bx)^2-2ax
p'(x) = (1-bx+bxRT)/(1-bx)^2-2ax

Answer 44

\[\frac{ RTx }{ 1-bx }\] this is the first part right?
since RT is a constant we can take it out just from this term
\[\rm RT(\frac{ x }{ 1-bx })\]
now you can take the derivative of the fraction with respect to x (ignoring the constant)
but let me just do the way you were doing it
\[\rm P(x)= \frac{ \color{red}{RTx} }{\color{blue}{ 1-bx }}-ax^2\]
\[\rm P(x)' =\frac{ d }{ dx }(\frac{ \color{red}{RTx} }{ \color{blue}{1-bx}})-\frac{ d }{ dx }
(ax^2)\]
applying the quotient rule for the first term
\[\rm \frac{ \color{red}{RT}\color{blue}{\color{blue}{(1-bx)} }-\color{red}{RTx}(\color{blue}{-b})}{ \color{blue}{(1-bx)^2} }\]
simplify
\[\rm \frac{ RT-RTbx +RTbx }{ (1-bx)^2}\] \[\rm \frac{ RT\cancel{-RTbx +RTbx }}{ (1-bx)^2}=\frac{RT}{(1-bx)^2}\]same answer right?
but looks like too messy so its better to take out the constant first (at least thats what i like to do)if you are comfortable doing like this
go for it but make sure you use parenthesis to avoid the mistakes

Answer 45

So that would be the derivative?

Answer 46

p'(x) = (RT)/(1-bx)^2 -2ax

Answer 47

@Hero

Answer 48

\(\color{#0cbb34}{\text{Originally Posted by}}\) @AbuZZ
p'(x) = (RT)/(1-bx)^2 -2ax
\(\color{#0cbb34}{\text{End of Quote}}\)
yes that's right.

Answer 49

So how would I do part C?

Answer 50

for part C it is asking to find P(x)' when x=0

Answer 51

Linear Approximation formula:
y=f(a)+f'(a)(x-a)
Plugging in P(x) we get 0/1 = 0
y = 0 + RT(x-0)
y = RTx
Is that correct?

Answer 52

Note, I got RT by plugging in 0 in the derivative we found.
P'(0) = RT(1-bx)^2-2ax
P'(0) = RT(1-0)^2-2a(0)
P'(0) = RT - 0
P'(0) = RT

Answer 53

is that formula given??

Answer 54

No, but I just learnt it. And I think I just figured out how to do the rest of the problem. Thank you so much for your help, I truly appreciate it. I might have some questions in the near future though. :P

Answer 55

RTx is correct.
np
sorry i'm doing something else that's why i am getting distracted and not being able to response on time.
i am looking forward to do some more calc!!

Answer 56

It's fine, your slow response made me figure it out by myself which is always better! Inshallah. ((:

Answer 57

mashALLAH !! :o

Answer 58

https://i.gyazo.com/3374fe99db51c7eea9b6cdfbcf239a76.png
https://gyazo.com/d8c51d5501d9db013119c35641b53ae3

Answer 59

Never mind figured that out. Lol, ty for all your help tho.