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

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) #