Question

Find dw/dt by using the chain rule approximation and convert w to a function of t before differentiating.

w=cos(x-y),x=t^2,y=1

Answer 1

I just want to know if I'm doing this right:

\[dw/dt(dx.dt)+dw/dy(dy/dt)=-\sin(x-y)(2t)-0\]
Plugging in both x and y:

\[-\sin(t ^{2}-1)(2t)\]

Answer 2

oh and from there my answer for dw/dt is -2tsin(t^2-1)

Answer 3

Now doing the second part I get:

\[w=\cos(t^2-1)=-2tsin(t^2-1)\] is this right?

Answer 4

\[dw = (dw/dx)dx + (dw/dy)dy = -sen(x-y)dx +sen((x-y)dy\]
that would be de dw
now: dx = 2t, dy = 0
pluging in:
\[-sen(t ^{2}-1)2t\]
this would be dw/dt

Answer 5

ok thanks.