Solving Treg Differentiation Questions with -1 Power

[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

 

[rl.bhat]

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

 

[HallsofIvy]

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 ?

Yes, that is 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

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".

 

[Mspike6]

Thankv you guys

So it will be

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

right?