Differentiate with respect to x

[lubo]

Homework Statement

Differentiate -sin²x

Homework Equations

CHAIN RULE dy/dx=dz/dx*dy/dz
PRODUCT RULE

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.

 

[CompuChip]

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

 

[lubo]

(-sin x)² is not the same as - (sin² x).].

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 you let z = -sin x, then y = -z².

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.

 

[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.

 

[lubo]

Excellent, many thanks.