Question

Question #c2fda

Answer 1

`f'(x)=-pi/6sin(pit/6)`

Explanation:

So here we have a trig function and whenever we see trig functions, the first thought that should pop into our head is that we need to use the chain rule. So to start, we will take the derivative of the outside, which in this case would be the `cos()` around the outside which gives us `-sin(pit/6)`.

However, the chain rule then requires we multiply this derivative by the derivative of the inside part of the function, `pit/6` which would give us `pi/6` since the derivative of `t=1` which gets multiplied by the constant `pi/6` leaving us with just that. When multiplied, our answer looks like `f'(x)=-pi/6sin(pit/6)`.

Answer 2

`v(t) = -sin(pi*t/6)*(pi/6)`

Explanation:

The derivative of cosine is negative sine. So
`f'(t)=-sin(pi*t/6)`
But according to the chain rule, you have to multiply this by the derivative of the contents inside the parentheses.
The derivative of `pi*t/6` with respect to t is `pi/6`.
So the velocity function would look like:
`f'(t) = -sin(pi*t/6)*(pi/6)`
`v(t) = -sin(pi*t/6)*(pi/6)`