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Answer 1 · 2017-05-16T19:07:24

Using #f'(1) = lim_(xrarr1) (f(x)-f(1))/(x-1)#, I get #f'(1) = 10#

#f'(1) = lim_(xrarr1) (f(x)-f(1))/(x-1)#

# = lim_(xrarr1) (e^(x^10-1)+(x-1)^2sin(1/(x-1))-1)/(x-1)#

# = lim_(xrarr1) (e^(x^10-1)-1)/(x-1) + lim_(xrarr1) ((x-1)^2sin(1/(x-1)))/(x-1)#

Now, # lim_(xrarr1) (e^(x^10-1)-1)/(x-1) # is #g'(1)# for #g(x)=e^(x^10-1)#.

And, #g'(x) = 10x^9e^(x^10-1)#, so #g'(1) = 10#

The second limit,

#lim_(xrarr1) ((x-1)^2sin(1/(x-1)))/(x-1) = lim_(xrarr1) ((x-1)sin(1/(x-1))) = 0# by the squeeze theorem.

Therefore,

#f'(1) = 10#

Interestingly, the derivative exists at #1#, but is not continuous at #1# because the limit as #x# approaches #1# of #f'(x)# fails to exist.

#f'(x) ={(10,"if",x=1), (10x^9e^(x^10-1)+2(x-1)sin(1/(x-1))-cos(1/(x-1)),"if",x != 1):}#