Melody

Sorry 

I am on the road today.

It is nice to be wanted/needed though.

Thank you.  :)

  atan is the same thing.  :)

Just enter it straight into an calc.

on the site calc 8C5 is entered as    nCr(8,5)

Oh ok   Thanks  

I have not seen capital pi used like that before.    

Is Geno's answer helpful ?

$$\\y'=(x+1)* 8(2x-3)^3+(2x-3)^4\\
y'=8(x+1)* (2x-3)^3+(2x-3)^3*(2x-3)\\
y'=(8x-8)* (2x-3)^3+(2x-3)^3*(2x-3)\\
y'=(2x-3)^3[(8x-8)+(2x-3)]\\
y'=(2x-3)^3[8x+2x-8-3]\\
y'=(2x-3)^3[10x-11]\\
y'=(10x-11)(2x-3)^3\\$$

Look what I found

I am all excited!

Estoy muy emocionada!

$$y = (x + 1)(2x-3)^4$$

for this one you will use the product rule

y=uv 

y'=uv'+vu'

u=x+1                      u'=1

v=(2x-3)^4               v'=[4(2x-3)^3]*2=8(2x-3)^3

and from here you should be able to finish it    

$$\\y=\frac{ln(x+2)}{x+2}$$

use quotient rule

$$\\If \;\;y=\frac{u}{v}\\\\
y'=\frac{vu'-uv'}{v^2}\\\\\\
u=ln(x+2)\\
u'=\frac{1}{x+2}\\\\
v=x+2\\
v'=1\\\\\\$$

Can you finish it?

What kind of a graph do you want to make?