Question

If `f(x) = cosx * cos2x * cos4x * cos8x * cos16x` then `f'(pi/4)` is ??

Answer 1

`f'(pi/4) = sqrt(2)`

Explanation:

We have:

`f(x) = cosx * cos2x * cos4x * cos8x * cos16x`

We could use the product rule with `5` products and form a full equation for the derivative, but this is not required, as we just need the value of the derivative when `x=pi/4`.

We can save a lot of unnecessary computation by using the following:

`x=pi/4 => { (cosx,sqrt(2)/2), (cos2x,0), (cos4x,-1), (cos8x,1), (cos16x,1) :} \ \ \ \ \` and `{ (sinx,sqrt(2)/2), (sin2x,1), (sin4x,0), (sin8x,0), (sin16x,0) :}`

And when we apply the product rule we will get a function of the form:

`f'(x) = (cosx)' \ cos2x \ cos4x \ cos8x \ cos16x +`
`" " cosx \ (cos2x)' \ cos4x \ cos8x \ cos16x +`
`" " cosx \ cos2x \ (cos4x)' \ cos8x \ cos16x +`
`" " cosx \ cos2x \ cos4x \ (cos8x)' \ cos16x +`
`" " cosx \ cos2x \ cos4x \ cos8x \ (cos16x)'`

And so with `x=pi/4` any term containing the derivative of `cos4x, cos8x, cos16x` will depend upon `sin4x, sin8x, sin16x` all of which are zero. Similarly any term containing `cos2x` will be zero.

So the only non-zero terms of consequence are:

`{: f'(x) ]_(x=pi/6) = cosx \ (cos2x)' \ cos4x \ cos8x \ cos16x`

`" " = cosx \ (-2sin2x) \ cos4x \ cos8x \ cos16x`

`" " = sqrt(2)/2 * -2 * -1 * 1 * 1`
`" " = sqrt(2)`

Hence `f'(pi/4) = sqrt(2)`