1. ## Differentiation

12.5 sin(0.06 t)

What rule would I use to differentiate it?

2. ## Re: Differentiation

Originally Posted by Thepiman1
12.5 sin(0.06 t)

What rule would I use to differentiate it?
I presume you know what the differential of the sine function is. Is that correct?

Now you use a substitution and the chain rule.

$y = 12.5sin(0.06t).$

$u = 0.06t \implies \dfrac{du}{dx} = what?$

$Also\ u = 0.06t \implies y = 12.5sin(u) \implies \dfrac{dy}{du} = what?$

How do you calculate $\dfrac{dy}{dx}$ knowing $\dfrac{du}{dx}$ and $\dfrac{dy}{du}$?

3. ## Re: Differentiation

du/dx= 0.06
dy/du= 12.5 cos (0.06)
dy/dx= 0.06 x 12.5 cos (0.06)

Look right?

4. ## Re: Differentiation

Originally Posted by Thepiman1
du/dx= 0.06
dy/du= 12.5 cos (0.06) No. 12.5cos(u) = 12.5cos(0.06t)
dy/dx= 0.06 x 12.5 cos (0.06)

Look right? NO, see below
First, I see that I made an error in my first post. I should have been using dt, not dx. That may have confused you. If so, I apologize.

$y = 12.5sin(0.06t).$

$Let\ u = 0.06t \implies \dfrac{du}{dt} = 0.06.$ You were right here.

$Also\ u = 0.06t \implies y = 12.5sin(u) \implies \dfrac{dy}{du} = 12.5cos(u) = 12.5cos(0.06t).$

$\therefore \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dt} = 0.06 * 12.5cos(0.06t) = 0.75cos(0.06t).$

It is not technically incorrect to leave an arithmetic expression in an answer, but if the arithmetic expression is easily computable, I think it is good form to do the computation. So I would not give an answer of 0.06 * 12.5cos(0.06t).

EDIT: A lot of the mechanics in differential calculus is just using substitutions to reduce a complex expression into one or more simple expressions that have a derivative you have memorized and then using the chain rule. The formal step of doing the substitutions explicitly is often skipped to save time (although it increases the risk of errors), but whenever you see something really complex or unknown to you, think about explicitly trying one or more substitutions.