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## #1 2006-04-25 12:33:49

silvercity87
Member
Registered: 2006-01-28
Posts: 5

### Finding the derivative

How do I find the derivative of

y^2=sin^4(2x)+cos^4(2x) ?

I know that for y^2, i have to use implicit differentiation.

But for sin^4(2x), is it like saying (sin^2(2x))^2??

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## #2 2006-04-25 14:19:44

George,Y
Member
Registered: 2006-03-12
Posts: 1,319

### Re: Finding the derivative

(sin^2(2x))^2-That's just one way.

[sin²(2x)²]'=2sin²(2x) [sin²(2x)]'=2sin²(2x) 2sin(2x)[sin(2x)]'

=4sin³(2x)cos(2x)[2x]'=4sin³(2x)cos(2x)2=8sin³(2x)cos(2x)

Last edited by George,Y (2006-04-25 14:20:30)

X'(y-Xβ)=0

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## #3 2006-04-25 14:41:20

silvercity87
Member
Registered: 2006-01-28
Posts: 5

### Re: Finding the derivative

oh. ok.
so for cos^4(2x)
the derivative would be
[cos²(2x)²]'=[2cos²(2x)]*[2cos(2x)]*[-sin(2x)][2]
=-8cos³(2x)sin(2x)

and the whole equation would be,

2y (dy/dx) = 8sin³(2x)cos(2x) - 8cos³(2x)sin(2x)
dy/dx  = {8sin(2x)cos(2x)[sin²(2x)-cos²(2x)]} / 2y

... but how do i go further?
i am supposed to make the dy/dx = (-sin8x)/y

thanx a lot for the answer

Last edited by silvercity87 (2006-04-25 14:41:57)

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## #4 2006-04-27 02:54:17

George,Y
Member
Registered: 2006-03-12
Posts: 1,319

### Re: Finding the derivative

It depends on whether y is defined as nonnegtive or nonpositive. In either case, you can solve y explicitly and substitude y with the expression in the solution, and get a global solution on a certain domain.

Or, you already know the point (x[sub]0[/sub],y[sub]0[/sub]), and you want to find the tangent slope at it, you will easily get a local solution.

By the way,

d(a[sup]k[/sup])/dk = k a[sup]k-1[/sup]

Last edited by George,Y (2006-04-27 02:55:42)

X'(y-Xβ)=0

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