Question

Question #ed8da

Answer 1

`d/dx(cos(x/2))=-1/2sin(x/2)`

Explanation:

This derivative requires knowing the derivative of cosine and knowing the chain rule:
`d/dx(cos(x/2))=-sin(x/2)*(1/2)=-1/2sin(x/2)`

Answer 2

Break it down into function compositions, then use the chain rule to solve for its derivative.

Explanation:

The function `cos(x/2)` first takes in an input `x`, multiplies it by `1/2`, then takes the cosine of that result. Or, to put it in another way:

`cos(x/2)` goes from `x rarr x/2 rarr cos(x/2)`

So, we can break this down into a function composition. In other words, we can break this down into several functions, each of which do only one step. In this case, we have two:

`f_1(x) = x/2`

`f_2(x) = cos(x)`

So `cos(x/2)` can be rewritten as `f_2(f_1(x))`. That might look confusing, but don't worry, it's just doing each step, from the inside, out. `x` is first passed on to `f_1` to multiply by half, then the result of that multiplication is passed on to `f_2`.

Now, to take the derivative, use the chain rule. What it tells us, is that we can use "intermediate functions" when taking the derivative.

In this case, this means that to take the overall derivative, we can take the derivative of the first function, then take the derivative of the second function with respect to the first function, before finally evaluating the first function. Here's what it looks like algebraically (where `f(x) = f_2(f_1(x))`):

`(df)/(dx) = (df_1)/(dx) * (df_2)/(df_1)`

Personally, I see the chain rule as more of a method than a rule. Here's how it goes:

First, find out how much `f_1` changes as `x` changes (by taking its derivative). To do that, take the derivative, as such:

`(df_1)/(dx) = d/dx (x/2) = 1/2 * d/dx x = 1/2 * 1 = 1/2`

What was done is that the constant was "thrown out" of the derivative, which can be done because the constant merely scales the derivative; and then the derivative of `x`, that is, the slope of a line `y = x`, is solved to be `1`, which is then multiplied to the constant to get a final result of `1/2`.

Notice, when we took this derivative, we divided a tiny nudge in `f_1`, denoted `df_1`, by a tiny nudge in `x`, denoted `dx`, where the letter `d` indicates that they are approaching zero. This means that we can somehow solve for `df_1` by "multiplying" `dx`:

`(df_1)/(dx) = 1/2 rarr df_1 = 1/2 dx`.

Now, we have enough information to take the derivative of the second function, `f_2`. We are, however, going to shove `f_1` as the input, at least for now.

`(df_2)/(df_1) = d/(df_1) cos(f_1) = -sin(f_1)`

"Multiplying" by `df_1`:

`df_2 = -sin(f_1) df_1`

And now, we can evaluate both `f_1` and `df_1`:

`df_2 = -sin(x/2) 1/2 dx`

Simplify:

`df_2 = -1/2 sin(x/2) dx`

And "divide" by `dx`:

`(df_2)/(dx) = -1/2 sin(x/2)`

This is not just the derivative of `f_2` alone, but because we used `f_1` as an intermediate function, it is the derivative of the function `cos(x/2)` as a whole:

`(d(cos(x/2)))/(dx) = -1/2 sin(x/2)`