If #f(x)=sin x# and #g(x)=e^x#, why is #f'(x)=cos x# and #g'(x)=e^x#?

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Answer 1 · 2017-05-21T14:45:09

#f'(x)=cosx#, iff #x# is measured in radians, and #g'(x)=e^x# by definition.

If we have a function, #f(x)#, then we can define the derivative of that function, #f'(x)# as:

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

Letting #f(x)=sinx#,

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

#=lim_(h->0)# #(sinxcos h+cosxsinh -sinx)/h#

#=lim_(h->0)# #((sinxcos h)/h) + lim_(h->0)# #((cosxsin h)/h)#
#-lim_(h->0)# #(sinx/h)#

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

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

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

#thereforesinx# #lim_(h->0)# #(cos h/h-1/h) + cosx# #lim_(h->0)# #(sin h/h)#

#=sinx xx0+cosx xx1#

#=cosx#

Note: If #x# is not measured in radians, then the evaluation of the limits is different and #(sinx)'=0.0175cosx#.

Applying the same limit to #g(x)#, we eventually get:

#g'(x)= e^x# #lim_(h->0)# #((e^h-1)/h)#

#lim_(h->0)# #((a^h-1)/h)=lna#

#therefore e^x# #lim_(h->0)# #((e^h-1)/h)#

#=e^xlne#

#=e^x#