P= a - bN {(p-t)/wc}^1/1-z wer z>1
differentiate tis expression wrt ' t'

- anonymous

P= a - bN {(p-t)/wc}^1/1-z wer z>1
differentiate tis expression wrt ' t'

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- RadEn

too many variables there :)

- anonymous

dat the prob!! can u solve tis its profit maximizing level of output inclusive of tax...had to be weird..

- anonymous

|dw:1356073901037:dw|

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## More answers

- Kainui

\[P=a-bN (\frac{ p-t }{ wc })^{(\frac{ 1 }{ 1-z })}\] differentiate with respect to t when z>1 right? I think we can do this.

- anonymous

@Kainui plz do

- Kainui

It seems confusing but try to take it one bit at a time. Remember that all of these are essentially constants like the number 5, 23, or whatever. Give it your best shot and show me what you get.

- Kainui

You should try splitting up the fraction within so that it's more easy to look at. I suggest trying to isolate "t" because everything else is just a constant and will lead to confusion.

- anonymous

i've tried n ... m getting tha same crap .. can u show me ur steps plz!!

- shubhamsrg

i have never knows about taxes much,,just confirming..
we have to find DP/dt right ? given t is the only variable and all others are constants ?

- shubhamsrg

known*

- anonymous

@shubhamsrg dats rite u dont need to knw abt taxes or nething its jst a typical expression!

- Kainui

Alright I'll write out some steps because this is particularly confusing, give me a moment to make it clear in my description.

- shubhamsrg

i see..
then check this please :
|dw:1356074934050:dw|

- anonymous

dats exactly wat m getting but the soln given has a numerator and denomenator

- shubhamsrg

whats the solution ?

- Kainui

\[P=a-bN( \frac{ p-t }{ wc })^{\frac{ 1 }{ 1-z }}\]
Now since everything else is constant, separate out the t into its own term to make it clear what we're differentiating:
\[P=a-bN( \frac{ p }{ wc }-\frac{ t }{ wc })^{\frac{ 1 }{ 1-z }}\]
Now from here you can take the derivative more simply by using the chain rule.
You'll see that a goes to 0, since the derivative of any constant is 0. We're doing this with respect to t.
Now if you use the chain rule on the second part, let's consider a simpler looking expression:
\[5(3-2t^2)^7\]
this looks exactly like the second part of the problem, yes? When we use the chain rule we take the derivative of the outside and multiply it by the derivative of the inside like this:
\[35(3-2t^2)^6* (-4t)\]
You must do the same thing here.

- Kainui

That looks like what I got too, I even simplified it down to the same exact thing as you.

- anonymous

hey m sorry but there is minor change in ques... instead of w its z m sorry guy n lemme post dat ans for

- Kainui

Not much will change, suneja since both are constants.

- shubhamsrg

yep..exactly!

- Kainui

Don't get caught up in looking at them as letters. Do the exact same thing you would do as if they were just any other constant number.

- anonymous

|dw:1356075429814:dw|

- anonymous

@shubhamsrg @Kainui this is the ans .... i got the same ans as u guys
now can u guys show me how did they get tis one and mind u its NOT WRONG ANS as its has economic interpretation attached

- Kainui

Hmm well what's the economic interpretation? I suppose the only difference is just some algebraic manipulation?

- anonymous

here denum is = 1+numerator so dp/dt <1

- Kainui

So wait, you have to get it in the same form as this? I'm not sure I understand what you want here.

- anonymous

@Kainui jst dP/dt dat et

- Kainui

I think you got it.

- anonymous

guys thanks for ur tym n effort... but i guess i'l hav to figure it out maslf!!

- Kainui

Good luck!

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