# Differentiate with respect to x

1. Jan 4, 2012

### lubo

1. The problem statement, all variables and given/known data

Differentiate -sin2x

2. Relevant equations
CHAIN RULE dy/dx=dz/dx*dy/dz
PRODUCT RULE

3. The attempt at a solution

Using the chain rule;

let z = -sinx let y = z2

dz/dx = -cosx dy/dz = 2z

dy/dx=dz/dx*dy/dz = -cosx*2-sinx = 2cosx*sinx

If I use the chain rule:

let u = -sinx let v = -sinx

du/dx = -cosx dv/dx = -cosx

udv/dx+vdu/dx = -sinx*-cosx + -sinx*-cosx

= 2sinx*cosx

My book answer says -2sinxcosx ?

Also on a website says:

Expression

Function -sin(x)^2
f'x = -2*cos(x)*sin(x)

Thanks for any help.

2. Jan 4, 2012

### CompuChip

(-sin x)2 is not the same as - (sin2 x).
If you let z = -sin x, then y = -z2.

3. Jan 4, 2012

### lubo

What about -sinx*sinx. Does this equal -sinx2 or -sin2x ? or what you have put?

I would write (-sin x)2 = (-sinx)(-sinx) ? (but this = sinx2 and thus it might be -(sinx*sinx) ?and you are saying it is not equal to - (sin2 x) = -sinx*sinx. I think?

Could you put a few lines together to explain this please?

Why did you put the - sign outside of the brackets?

If I do it like you said I get:

let z = -sinx let y = -z2

dz/dx = -cosx dy/dz = -2z

dy/dx=dz/dx*dy/dz = -cosx*-2-sinx = -2cosx*sinx

I thought that the minus sign would be taken after within Z and it would not need to be added outside as in the first instance?

Thanks.

4. Jan 4, 2012

### CompuChip

Squaring has higher priority than multiplication (by -1), so when we write -z2, we mean -(z2) = - z * z.
This is something else than (-z)2, which would be (-z) * (-z) = z * z = z2.

In the case of a sine function, we abbreviate (sin x)2 to sin2 x. This is done to distinguish from sin x2 = sin(x2) while not having to write the brackets.

So -sin2 x means negative the square of the sine of x: -1 * (sin x) * (sin x), as opposed to -sin(x2) (which cannot be simplified further) or (-sin x)2 = (-sin x) * (-sin x) = sin x * sin x = sin2 x.

5. Jan 4, 2012

### lubo

Excellent, many thanks.