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That makes sense - thanks Alan.    

Thanks very much Alan.  

Alan, I just drew these graphs.  I think that they are ok but could you check them please.

Unfortunately the pictures do not seem to be displaying properly tonight.  Oh well maybe it will appear properly tomorrow.  

Here is an online version.  It is the same.

I think he/she just wants the expression evaluated for x = -2

7 - 4*(-2) + 2*(5*(-2) + 8)

7 - (-8) + 2*(-10 + 8)

7 + 8 + 2*(-2)

7 + 8 - 4

11

Stu, why have you taken this attitude.  

This series of posts have been really fun.

I thought you were starting to go with the flow.  Your recent rebuttle was really clever and funny.

My oppinion of you rose markedly.  Now I am dashed.   

We don't want you to leave the forum.  Why don't you lighten up and enjoy the fun.  

1 + 3 = 7.58

Stu and Archy

$$\int \frac{f'(x)}{f(x)}dx=ln[f(x)]+c$$

This is an integral that you need to automatically recognise!

I'm sorry there really is no question here.

Dragon, this can be taken a little further

$$a(2*x)=a*2*x=2*a*x=2ax$$

i saw that on the formulas and thanks.

To the mods: if these posts will not be removed from being of topic and derailing i will find a forum which is better moderated. now i will stop rebuting this t**d's post and leave you all to do the job of moderating. Thanks.

In the case of the integral you use the product form the other way round

Starting with  d(u*v)/dx = u*dv/dx + v*du/dx

Rewrite as  u*dv/dx = d((u*v)/dx - v*du/dx

Integrate each term 

∫udv = u*v - ∫vdu

Wow, nice try. What a ridic last post anon. You are so out if your depth and so lonely you should really go try joining some of your kind in zoo enclosure near you. Now stfu and gtfo of this forum, back to your honest posting as a member before I out you!

This will involve the sin function.

The vertical displacement will be 10*(sin(60°) - sin(30°))m  (I'm assuming the angle is between the boom and the horizontal).

sin(60°) = (√3)/2  and sin(30°) = 1/2

I'll leave you to complete the calculation.

but they cannot solve for integal?

Both product and quotient rule are for differentiating a product.  It just happens that the "product" involves a factor in the denominator when using the quotient rule.

In fact, I never bother trying to remember the quotient rule.  I just use the product rule combined with the chain rule to differentiate a quotient.  So, for example I would differentiate y = u/v as follows

y = u*v^-1

dy/dx = u*d(v^-1)/dx + v^-1*du/dx

dy/dx = u*d(v^-1)/dv*dv/dx + v^-1*du/dx

dy/dx = -(uv^-2)*dv/dx + v^-1*du/dx

dy/dx = -(u/v²)*dv/dx +(1/v)*du/dx

Of course this can be rearranged to look like the standard quotient formula, but I find it easier just to apply the steps above directly to the specific functions involved!

solved. thanks for the assit.

i was going to go with substitution method haha, i still get confused by making the term dy but a little ok i guess. would the product rule have worked or not worked?

Alan would it be a product rule or quotient rule? One is for integral and one for derivatives, opposites to each other?

It can be done more simply Stu.

Let y = x² + 1

Then dy = 2xdx

∫2x/(x²+1)dx = ∫dy/y = ln(y) = ln(x² + 1)  (plus an arbitrary constant of integration of course).

wrong and removed

According to "The Hitchhiker's Guide to the Galaxy" "The answer to the ultimate question of life, the universe and everything" is 42. 

I know one thing, though: the meaing of life isn't to insult people..

(-4/-1)  +  (-4/3)  -  (-1/3)

Do the terms I've put in brackets first:

4  -  4/3  +  1/3       because if you divide one negative number by another you get a positive number; a plus followed by a minus is a minus; and a minus followed by a minus is a plus.

Now put every term over a common denominator.  Since the last two terms are both over 3, express the first term as 12/3  (since 12/3 = 4). 

12/3  -  4/3  +  1/3

Now we can simply write

(12 - 4 + 1)/3  or 

9/3  or just 3

That is correct Chris!

There is also enough information to determine that the initial horizontal velocity is 32.9/2.27 ≈ 14.5 m/s.