Solving Differential Equations

[mattmannmf]

Solve the following:

d/dt cos(theta)
d/dt t sin(theta)
d/dt r cos (theta)
d/dt r^2 (theta)
d/dt e^ (-3x)
d/dt (x^2 + y^2)

I would assume all by the second one are 0 since your solving for terms dt and not theta, x, y, or r... I don't think its right at all. I know it goes something like this:
d/dt f(x) = dy/dx * dx/dt
I just am not sure how to grasp what I'm doing wrong.

 

[hunt_mat]

Is theta a function of t?

 

[mattmannmf]

what do you mean?

 

[hunt_mat]

is \theta =\theta (t), otherwise the derivative will be non-zero.

 

[mattmannmf]

all it says its differential calculus and gives the problem as I stated above

 

[Mark44]

From the title of the thread ("Calculus Chain Rule"), I think it's reasonable to assume that \theta is a differentiable function of t, and that you are meant to use the chain rule.