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1 Answer
May 16, 2017

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

Explanation:

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):}