What does this question mean exactly?

I have this problem

The question states evaluate at dQ/dt at t=3 for Q =
(3x^3 + 2x^2 - 9)/x^2 where x is not equal to 0

I have no idea what this question means, do I differentiate it all and that's it??

I ended up doing this (9x^2 + 4x)/2x

Is this right is this answering the question

Reply 1

Notnek

There must be more to the question that that. Are you sure you've posted everything?

Reply 2

james.h

As notnek has said, you don't appear to have posted the entire question. How does x relate to t?

I imagine the procedure would be:

Using the chain rule, find

\frac{\text{d}Q}{\text{d}t} = \frac{\text{d}Q}{\text{d}x} \frac{\text{d}x}{\text{d}t}

as a function of t.

Then substitute the value t = 3 into this expression to get the answer.

(edited 11 years ago)

Reply 3

BabyMaths

Original post by mox123

Simplify first

Q=\frac{3x^3}{x^2}+\frac{2x^2}{x^2}-\frac{9}{x^2}=??

Edit: good point about the x and t guys. Bad reading on my part, but my advice may still be useful.

(edited 11 years ago)

Reply 4

mox123

This is all there is, it is a differentiation question from what I can see

2012-06-15 17.04.26.jpg9.3KB

Reply 5

mox123

I did simplify it, this is what I get, is this the right answer?

2012-06-15 17.06.57.jpg21.1KB

Reply 6

BabyMaths

Original post by mox123

This is all there is, it is a differentiation question from what I can see

Then I think they meant evaluate dQ/dx at x=3.

Reply 7

mox123

the problem is I've not been taught the chain rule yet and the book doesnt teach this either, I've to use another method and I'm not sure if the method I've used is correct?

yeah I think maybe it's a typo, x=3 instead of t=3

is my working correct in the above attached images?

Reply 8

BabyMaths

Original post by mox123

I did simplify it, this is what I get, is this the right answer?

Yes. There's an error along the way but you have the right answer in the end. Sub in x=3..if we've interpreted the question correctly.

Reply 9

mox123

ok so substitute x=3 into it then complete the equation? is that my error?

Reply 10

mox123

i think I got the answer right 3 + 18/x^3 is this correct?

Reply 11

james.h

Looks like a typo. :holmes:

dQ/dx it is, then.

Yes, dQ/dx = 3 + 18/x³. Then just substitute in x = 3 and you're done.

The issue with your working is that you keep saying everything's equal. The original function Q is not identical to its derivative, dQ/dx! The working would be correctly expressed as:

\displaystyle \frac{\text{d}}{\text{d}x}\left( \frac{3x^3 + 2x^2 - 9}{x^2} \right) = \frac{\text{d}}{\text{d}x}\left( 3x + 2 - 9x^{-2} \right) = 3 - (-2)(9)x^{-2-1} = 3 + 18x^{-3}

(edited 11 years ago)

Reply 12

mox123

ahh right, sorry I didn't know, I've not been taught properly i'm afraid and trying to pick things up as I go along :-( textbook that I've been given is full of errors, I'm also using Maple and Mathematica as well as Bagatrix Calculus to keep me on track

Reply 13

Notnek

Original post by mox123

This is all there is, it is a differentiation question from what I can see

That's a pretty bad typo. Are you sure there's nothing else important above or below the section you posted?

Reply 14

james.h

Practise makes perfect. :biggrin:

You may find this a useful resource for quickly checking your answers in future. :yep:

EDIT: notnek makes a valid point...though the OP did mention they hadn't covered the chain rule yet?

(edited 11 years ago)

Reply 15

mox123

yeh the book is full of mistakes as well as wrong answers in the back (i know its a bit of a shambles) which have been confirmed. I will be complaining to the school authority regarding alot of these mistakes. Simple things like not being told to use radians on your calculator to the above said.

I appreciate everyone making the effort to help here. Thanks again.

Reply 16

mox123

the chain rule has not been covered yet and will be covered next year, so the chain rule is out of consideration just now

Reply 17

mox123

anyone be able to help me with this?

the equation in question is 3sin(x+3)=cos(2x) to 4 dp between 0 =< x =< 7

by rearranging the equation i get f(x) = 3sin(x+3)-cos(2x) which im sure is correct as this is what i get in maple and calculus bagatirx

next step is to check by basically guess

so I guess 3.5 which is -0.10854 and 3.75 is 1.0034 so i think ok jackpot I'm close enough

anyway I differentiate it to 3COS(x+3) + 2sin(2x)

I then calculator using general iteration formula which is xnew= x(-0.10854) - f(x)/f'(x)

I run this continually until i get -0.1765 and this is what the newton-raphson calculator online confirms it as

my book states the answer is 3.524 and 6.1067 I MEAN WTF IS THAT??!! is my answer wrong, am I doing something wrong?

or is my book full of **** (which it has been lately, alot of stuff is incorrect but I can't rule it out that I haven't made a mistake or whether the fact the book is wrong)