Is f(x) differentiable at x=1?

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Given
http://www.mathhelpforum.com/math-help/vlatex/pics/105_fde5ac6b051b4fac473487c7b4afa9e5.png

Is f(x) differentiable at x=1?

I know that we have to prove
http://www.mathhelpforum.com/math-help/vlatex/pics/65_6fae3c52eaa96aaafdf2c225a900ea48.png

exist/does not exist at x=1. But how do I begin? It's a piece-wise function.

Thanks for your help.
 
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Start by finding out the reasons that limit might not exist. This is a common type of problem, so you should find examples in your textbook.
 
The idea is one sided limits. You can compute the derivatives of f for x>1, and x<=1, for the derivative f' to exist, f' must be continuous there. So you've reduced the problem to continuity.
 
To add to hunt mat's response, while a derivative function is not necessarily continuous, it is continuous where ever it is defined. That is why taking the left and right side limits of the derivative works- if they are the same, the function is differentiable and the joint value is the derivative at that point.

What is the derivative of x^2? What is its value at x= 1? What is the derivative of (x+ 1)/2? what is its value at x= 1?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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