Differentiate #sin(x/2)# using first principles?

Question

Isaiah Allen · Accepted Answer

# d/dx sin(x/2)=1/2cos(x/2)#

Let: # f(x) = sin(x/2) # By definition, the derivative of #f(x)# is the limit: # f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h # So for the given function, we have; # f'(x) = lim_(h rarr 0) ( sin((x+h)/2) - sin (x/2) ) / h # # " " = lim_(h rarr 0) ( sin(x/2+h/2) - sin (x/2) ) / h # Applying the trigonometric identity: # sin (A+B)=sinAcosB+sinBcosA # We get # f'(x ) =lim_(h rarr 0) ( sin(x/2)cos(h/2)+sin(h/2)cos(x/2) - sin(x/2) ) / h # # " " = lim_(h rarr 0) ( sin(x/2)(cos (h/2)-1)+sin(h/2)cos(x/2) ) / h # # " " = lim_(h rarr 0) ( (sin(x/2)(cos (h/2)-1))/h+(sin(h/2)cos(x/2)) / h )# # " " = lim_(h rarr 0) (sin(x/2)(cos(h/2)-1))/h+lim_(h rarr 0)(sin (h/2)cos(x/2)) / h# # " " = sin(x/2) \ lim_(h rarr 0) (cos (h/2)-1)/h + cos(x/2) \ lim_(h rarr 0)(sin (h/2)) / h# We know have to rely on some standard limits: # lim_(theta rarr 0)sin theta/theta =1 # # lim_(h rarr 0)(cos theta-1)/h =0 # And so manipulating the denominator by a factor of #2# to match the sine and cosine variable we have: # f'(x) = sin(x/2) \ lim_(h rarr 0) (1/2)(cos (h/2)-1)/(h/2) + cos(x/2) \ lim_(h rarr 0)((1/2)sin (h/2)) / (h/2)# # " " = 1/2sin(x/2) \ lim_(h rarr 0) (cos (h/2)-1)/(h/2) + 1/2cos(x/2) \ lim_(h rarr 0)(sin (h/2)) / (h/2)# And putting #theta=h/2# and noting that #h/2 rarr 0# as #h rarr 0# we have: # f'(x) = 1/2sin(x/2) \ lim_(theta rarr 0) (cos theta-1)/(theta) + 1/2cos(x/2) \ lim_(theta rarr 0)(sin theta) / (theta)# # " " = 1/2sin(x/2) xx 0 + 1/2cos(x/2) xx 1# # " " = 1/2cos(x/2)# Hence, # d/dx sin(x/2)=1/2cos(x/2)#

Charles Arocha · Answer

undefined To differentiate sin(x/2) using first principles, we'll apply the definition of the derivative:

f'(x) = lim(h->0) [f(x + h) - f(x)] / h

For sin(x/2):

f(x) = sin(x/2)

f'(x) = lim(h->0) [sin((x + h)/2) - sin(x/2)] / h

Using the angle sum identity for sine function, sin(a + b) = sin(a)cos(b) + cos(a)sin(b):

f'(x) = lim(h->0) [(sin(x/2)cos(h/2) + cos(x/2)sin(h/2)) - sin(x/2)] / h

Apply the limit:

f'(x) = lim(h->0) [sin(x/2)cos(h/2) + cos(x/2)sin(h/2) - sin(x/2)] / h

Using the trigonometric limit lim(h->0) (sin(h)/h) = 1:

f'(x) = lim(h->0) [sin(x/2)cos(h/2) + cos(x/2)sin(h/2) - sin(x/2)] / h
      = sin(x/2)lim(h->0) (cos(h/2)/h) + cos(x/2)lim(h->0) (sin(h/2)/h) - lim(h->0) (sin(x/2)/h)

f'(x) = (1/2)sin(x/2) + (1/2)cos(x/2) - (1/2)sin(x/2)

Combine like terms:

f'(x) = (1/2)(cos(x/2) + sin(x/2) - sin(x/2))
      = (1/2)cos(x/2)

Therefore, the derivative of sin(x/2) with respect to x using first principles is (1/2)cos(x/2).