How do you find the derivative of #f(x) = cos(pi/2)x# using the chain rule?

1 Answer
Jan 25, 2016

Answer:

For the function formulated in the question, the derivative is

#f'(x) = 0#.

However, the question might have meant this derivative instead:

#f'(x) = - pi/2 sin(pi/2 x)#

Explanation:

If you meant the function the way that you had typed it, then #cos (pi/2)# is just a coefficient of #x# and

#cos(pi/2) = 0#.

Thus,

#f(x) = cos(pi/2) * x = 0 * x = 0#

You don't need the chain rule for this one since the derivative of #f(x) = 0# is #f'(x) = 0#.

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You might have also meant

#f(x) = cos(pi/2 x) #

instead.

In this case, you can apply the chain rule as follows:

#f(x) = cos(color(blue)(u)) " where " color(blue)(u) = pi/2 x#

The derivative of #f(x)# is the derivative of #cos u# multiplied with the derivative of #u#:

#f'(x) = [cos u]' * u'#

It holds

#[cos u]' = - sin u = - sin (pi/2 x)#

#[u]' = [pi/2 x ]' = pi/2#

Thus, your derivative is

#f'(x) = [cos u]' * u' = - pi/2 sin(pi/2 x)#