Using chain rule to obtain the derivative dz/dt

[catch22]

Homework Statement

Homework Equations

dz/dt = dz/dx⋅dx/dt + dz/dy⋅dy/dt

The Attempt at a Solution

[/B]
I am getting :

=[-sin(x+7y) ⋅ 10t] + [-sin(x+7y) ⋅ 7 ⋅ (-1/ t²)]

then changing x and y terms:

=[-sin((5t²)+7(1/t)) ⋅ 10t] + [-sin((5t²)+7(1/t)) ⋅ 7 ⋅ (-1/ t²)]

 

[Ray Vickson]

Homework Statement

View attachment 91274

Homework Equations

dz/dt = dz/dx⋅dx/dt + dz/dy⋅dy/dt

The Attempt at a Solution

[/B]
I am getting :

=[-sin(x+7y) ⋅ 10t] + [-sin(x+7y) ⋅ 7 ⋅ (-1/ t²)]

then changing x and y terms:

=[-sin((5t²)+7(1/t)) ⋅ 10t] + [-sin((5t²)+7(1/t)) ⋅ 7 ⋅ (-1/ t²)]

Do you have a question?

 

[catch22]

Do you have a question?

yes, sorry, I'm just verifying If I had done it correctly.
the answer my teacher posted was :

which really got me questioning myself.

 

[Samy_A]

It seems that you teacher has $x(t)=5t^4$