Melody

I do not know what you are asking

but

$$\\\frac{d}{dx}(\frac{x}{1}+x^2)\\\\
=\frac{d}{dx}(x+x^2)\\\\
=1+2x\\\\$$

what is the other half for?

Thanks Alan

Welcome to the forum Mataxa   

I have given you the final answer that you need but do you need more working.

I did almost all of it with online calculators. 

Thanks Rosala. :)

Thank you Rosala.  That makes me very happy.

Why do you like them?   Do you actually look at all the posts?

How much of this do you need to do by hand?

Well here is a relevant graph

Just to be sure,

are these 

$$\\x^6=y^4+18\\
and\\
y^6=x^4+18$$       ???

@@ End of Day Wrap       Sun 16/11/14        Sydney,   Australia         Time 8:30 pm             ♬

Hi Everybody,

It was a little quiet today - some people don't do maths on the weekend - go figure ?

Anyway we still had great answers from CPhill, BrittanyJ, jillybean, geno3141 and Will85237.  Plus LancelotLink left 315 some cute and furry offerings on loan. 

If you want to find posts by a particular member this is how it is done. Up the top right of your screen is a search box.  Jusy type in  by:LancelotLink   (or the member that you are after) then his/her posts will be retreived:)

Now for a few interest posts:

1) Partial fractions - I haven't seen these on the forum before.  CPhill and Melody  

Chris and I learned/revised how to do them.  They are needed for some integrations.  

This gave us a small sense of achievement.  :)

Mon 17/11/14

1) Let's end the day with a laugh  Thanks Roasla.

That just leaves one more

c)

$$\\f(x)=\frac{x^2-10}{(x-2)(x+1)}\\\\
f(x)=\frac{x^2-10}{x^2-x-1}\\\\$$

We need to do an algebraic (or synthetic) division

$$\begin{tabular}{ccccccc} &&&\;1&&& &&&&||&||&||& \\
x^2&-x&-1&\|\;x^2&0&-10\\
&&&\;x^2&-x&-1\\ &&&||&||&||&& &&&&+x&-9&\\

\end{tabular}$$

so

$$\\\frac{x^2-10}{(x-2)(x+1)}=1\;+\;\frac{x-9}{x^2-x-1}\\\\
\frac{x^2-10}{(x-2)(x+1)}=1\;+\;\frac{x-9}{(x-2)(x+1)}\\$$

Now let

$$\\\frac{x-9}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{B}{x+1}\\\\
\frac{x-9}{(x-2)(x+1)}=\frac{A(x+1)+B(x-2)}{(x-2)(x+1)}\\\\$$

Equating numerators we have

$$\\x-9=A(x+1)+B(x-2)\\\\
x-9=(A+B)x+(A-2B)\\\\
so\\
A+B=1\quad\rightarrow\quad A=1-B\\
-9=A-2B\\
-9=1-B-2B\\
-10=-3B\\
B=10/3\\\\
A=1-10/3\\
A=1-3/3-7/3\\
A=-7/3$$

therefore

$$\\f(x)=\frac{x^2-10}{(x-2)(x+1)}\\\\
f(x)=1\;+\;\frac{x-9}{(x-2)(x+1)}\\\\
f(x)=1\;-\;\frac{7/3}{x-2}+\frac{10/3}{x+1}\\\\$$