Help differentiating?

Question

T = 25 - 4cos pi(t-3)/12
Please show working out :)

Sindri, I'm actually studying for a test and doing extra questions outside of my textbook that I set for myself. So whilst I appeciate your help don't blindly accusse people of things that you don't know are true.

Answer 1

How to differentiate
T = 25 - 4 cos [π(t - 3)/12]

So there is the derivative
dT/dt = d/dt {25 - 4 cos [π(t - 3)/12]}

Since the derivative of a sum (or difference) is the sum (or difference) of the derivatives, you can separate the differentiating symbol
dT/dt = d/dt (25) - d/dt {4 cos [π(t - 3)/12]}

The derivative of a constant is zero, so the 25 part can be eliminated
dT/dt = - d/dt {4 cos [π(t - 3)/12]}

Since the derivative of a constant times a function is the same as the constant times the derivative of a function, then the 4 can be removed from the derivative symbol
dT/dt = -4 d/dt {cos [π(t - 3)/12]}

We can use the chain rule:
Let u = [π(t - 3)/12]

Substitute u to the dT/dt above
dT/dt = -4 d/dt {cos u}

Using the chain rule:
dT/dt = -4 d/du (cos u) du/dt

Then getting du/dt, distribute π/12 in u = π(t - 3)/12
u = π/12 t - π/4
Thus,
du/dt = π/12

Since the derivative of cos is -sin, and du/dt = π/12,
dT/dt = -4 (-sin u)(π/12)

Multiplying and simplifying
dT/dt = π/3 sin u

Substitute back u = [π(t - 3)/12],
dT/dt = π/3 sin [π(t - 3)/12]

Therefore, we have differentiated T,
Hope this helps.....^_^
^_^
^_^

Answer 2

25 - 4cos (pi * (t - 3) / 12) = (you can differentiate each part of the sum separately) = (25)' - (4cos (pi * (t - 3) / 12)' =
- 4 * (cos (pi * (t - 3) / 12))' = ok, you know how to differentiate cosx (or cos t) but this is not it, this is a composition of 2 functions - cos t and pi * (t - 3) / 12 so you use formula for differentiation of composition (f(g(x)))' = f ' (g(x)) * g ' (x)
= - 4 * ( - sin( pi * (t - 3) / 12))) * ( pi * 12 )
= (pi / 3) * (sin (pi * (t - 3) / 12)

Answer 3

You want us to show the working out so you can just blindly copy it?

When you differentiate, constants go to zero and cos goes to negative sin and whatever's being operated on by cos also gets differentiated but is pulled out the front.

Therefore, your answer is T' = 4*(pi/12)*sin [pi(t-3)/12] but can be simplified a bit.

Answer 4

T' = ( 25 - 4cos pi(t-3)/12 )'

= ( 25 )' - ( 4 cos pi(t - 3)/12 )'
(derivative of sum or difference)

= 0 - ( 4 cos pi(t - 3)/12 )'
(derivative of constant)

= - 4 ( cos pi(t - 3)/12 )'
(derivative of constant times something)

= - 4 ( - sin pi(t - 3)/12 ) ( pi (t - 3) / 12 )'
(derivative of cosine is minus sine)
(chain rule)

= (4 sin( pi (t - 3) / 12 ) ) (pi / 12)
(derivative of constant times something)
(chain rule)

= (pi / 3) sin( pi (t - 3) / 12 )