How do you find f'(-5) using the limit definition given #f (x) = 5 cos(x)#?

1 Answer
Jun 6, 2017

#f'(-5)=-4.79#

Explanation:

Recall the limit definition of a derivative is:

#f'(x)-=lim_(h->0)# #((f(x+h)-f(x))/h)#

#f'(x)=5# #lim_(h->0)# #((cos(x+h)-cosx)/h)#

#=5# #lim_(h->0)# #((cosxcos h-sinxsin h-cosx)/h)#

#=5# #lim_(h->0)# #((cosxcos h-cosx-sinxsin h)/h)#

#=5# #lim_(h->0)# #((cosx(cos h-1))/h -(sinxsin h)/h)#

#=5(cosx# #lim_(h->0)# #((cos h-1)/h) -sinx# #lim_(h->0)# #(sin h/h)# #)#

#lim_(h->0)# #sin h/h=1#

#lim_(h->0)# #(cos h-1)/h=0#

#therefore5(cosx# #lim_(h->0)# #((cos h-1)/h) -sinx# #lim_(h->0)# #(sin h/h)# #)#

#=-5sinx#

#f'(x)=-5sinx#

#f'(-5)=-5sin(-5)=-4.79#

Note: the derivative of trig functions can only be evaluated as above given that #x# is in radians.