# Treg. Defrentiation

1. Dec 6, 2009

### Mspike6

I got 2 questions

First:
y= Cos3(5x2-6)

Solution :
Y' = 3cos2(5x2-6)(-sin(5x2-6))(10x)
y'= 30xCos2(5x2-6)(-sin(5x2-6))

Is the correct ?

Second
Y=3sin4(2-x)-1

I don't understand what do i have to do with the -1 (the power to the bracket )
so i add it to the 4 (the power of sin ) ?

am not sure

Any help is appreciated

2. Dec 6, 2009

### rl.bhat

(2-x)^-1 = 1/(2-x) = u. So y = 3*sin^4(u)
Now what is the derivative of 1/(2-x)?

3. Dec 6, 2009

### HallsofIvy

Yes, that is correct.

That's a very peculiar notation. It would be better with an additional pair of parentheses:
$Y= 3 sin^2((2-x)^{-1})$
That is, it is the (2- x) that is taken to the -1 power, not "sin".

4. Dec 6, 2009

### Mspike6

Thankv you guys

So it will be

y' = 12 sin3[(2-x)-1] Cos[(2-x)-1](-1)(2-x)-2(-1)

right?