Note: This isn't graded work, it's the basics..

 

Use the Chain Rule to get

dw/dt when,

w = cos(x-y)

x = t^2

y = sin t

 

I know dw/dt = (dw/dx)(dx/dt)+(dw/dy)(dy/dt)

 

Thus far I have,

dw/dt = (-sin(x)-cos(y))(2t)+(cos(x)+sin(y))(cos(t))

Substituting in x and y..

dw/dt = (-sin(t^2)-cos(sint)(2t))+(cos(t^2)+sin(sint)(cost))

 

I feel like I have done something wrong (I don't have the solution to this problem)..but I know it doesn't look correct.

 

I'm thinking that perhaps I made a mistake with the derivatives and it should be something closer to..

dw/dt = ((cos(x))(2t))+((sin(y)(cost))

dw/dt = ((cos(t^2))(2t))+((sin(sint))(cost))

But again, I'm clueless

 

Any help is greatly appreciated

 

Any help is appreciated

Edited October 26, 200916 yr by Cyanide

The chain rule is [imath]\frac{d}{dx}y(u(x))=\frac{dy}{du}\frac{dy}{dx}[/imath] or, in function notation, [imath](f \circ g)' = (f'\circ g) \cdot g'[/imath]. There shouldn't be an addition there (there is one in the product rule - but you don't need that).

 

Try finding functions [imath]f[/imath] and [imath]g[/imath] so that you can write [imath]w := f\circ g (t) = f(g(t))[/imath] (depending on your preferred way of writing functions).

 

Tip: don't bother differentiating anything w.r.t. [imath]x[/imath] or [imath]y[/imath].

Edited October 26, 200916 yr by the tree
+tip

The chain rule is [imath]\frac{d}{dx}y(u(x))=\frac{dy}{du}\frac{dy}{dx}[/imath]

 

Just like to correct a simple error on the above post.

 

[imath]\frac{d}{dx}y(u(x))=\frac{dy}{du}\frac{du}{dx}[/imath]

 

& rewrite the expression of the last post in a more familiar notation.

 

[imath]\frac{d}{dx}f(g(x))={f'}{(g(x)) ^ . }{g'(x)}[/imath]

Edited October 30, 200916 yr by psychlone
to correct LaTeX code

eep, yeah, of course. I'm not a massive fan of mixing d's and and flicks in notation, but each to their own I guess.