Help finding the derivative of a sine function!?

Question

Here is the function: y = (41 m) cos[πt/(5 s)] + (38 m)

Any help finding the derivative would be greatly appreciated. Thanks

Answer 1

The derivative of the second term "(38 m)" is 0. The derivative of c*f(t) is c * df(t)/dt. Thus the solution is going to be of the form (41 m)*df(t)/dt.

The f in that case being "cos[πt/(5 s)]".

The derivative of cos(u) is "-sin(u)*du/dt".

So the solution will have the form "-(41 m)*sin[πt/(5 s)]*du/dt".

So we are getting close. The derivitive of t is 1. So the derivative of "πt/(5 s)" is "π/(5 s)".

So the final solution is "-π*(41 m)/(5 t)*cos[πt/(5 s)]".

Answer 2

This is actually quite simple. Let's do the computation using the notations used in calculus.

We know that the derivative of any function in the form of

y = a cos bx,    is

dy/dx = (-a sin bx)(d(bx)/dx) = -ab sin bx, by the chain rule.

Therefore, the derivative of the given function is as follows:

y = (41 m) cos[πt/(5 s)] + (38 m)

dy/dx = d((41 m) cos[πt/(5 s)] + (38 m))/dx

dy/dx = - 41(π/5) sin [πt/(5 s)] + 0

dy/dx = - 41(π/5) sin [πt/(5 s)].

Answer 3

(-41π/5)sin(πt/5) is the function of the derivative. I just used the chain rule.