Chain rule problem: proper method?

[stevensen]

Homework Statement

Use the chain rule, the derivative formula D_xsinu=cosuD_xu, together with the identities
cosx=sin($\pi$/2 -x) and sinx=cos($\pi$/2 -x)
to obtain the fomula for D_xcosx.

Homework Equations

Chain rule: dy/dx=dy/du$\cdot$du/dx

The Attempt at a Solution

For my second attempt (the website made me log in again when I clicked on "Preview Post" last time - that seems to happen a lot):
D_xcosx=D_xsin($\pi$/2 -x)
=cos($\pi$/2 -x)(-1)
=-sinx

This is the result that I expected, but the formula in the textbook says D_xcosu=-sinuD_xu. I realize that this formula contains "u"s and the required formula does not. I also realize that D_xx=1, which does account for the discrepancy between the answer that I got and the formula in the book, but I can't help wondering if it was by sheer happenstance that it turned out to work. I see no progression that could lead me to obtain the result that D_xcosx=-sinxD_xx.

I'm not sure if I'm employing the chain rule properly. As you can see, I only used one variable in my proof, but the question uses both "u" and "x." Is there some form of notation that is considered correct for this kind of proof? (If so, I'd love to know why. There seem to be too many ways of expressing the same idea: dy/dx, d/dx, D_x, f'(x), others?) Thanks for your input.

 

[lanedance]

your derivative is correct, the only difference is the book leave it in a general form, for a general function u(x)

notation should go with the context of the problem, or how you learn it in class I guess, but all those you mention can mean the same thing. \

Generally if there's any chance of confusion I try and be as explicit as possible
$$\frac{dy(x)}{dx}$$