# Verify that #f(x)# is differentiable at #x = 0#? #f(x) = x((e^(1//x) - 1)/(e^(1//x) + 1))#

##### 1 Answer

The function is NOT differentiable at

graph{x((e^(1/x) - 1)/(e^(1/x) + 1)) [-1.065, 1.0684, -0.533, 0.5335]}

There are in fact different slopes from either side of

Well, you can multiply by

#f(x) = x((1 - e^(-1//x))/(1 + e^(-1//x)))#

Then, plugging in **continuous** at

#f(0) = 0 * ((1 - e^(-1//0))/(1 + e^(-1//0)))#

#= 0 * 1 = 0#

Since we have successfully found a determinate form and evaluated it, this function is *continuous* at

Now, we could very well have a function that has a **corner** or **cusp** at

The derivative is:

#f'(x) = x cdot (((1 + e^(-1//x)) cdot e^(-1//x) cdot 1/x^2 - (1 - e^(-1//x))cdot(-e^(-1//x) cdot 1/x^2))/(1 + e^(-1//x))^2) + (1 - e^(-1//x))/(1 + e^(-1//x))#

Simplify by factoring out

#= (-2e^(-1//x))/(x(1 + e^(-1//x))^2) + (1 - e^(-1//x))/(1 + e^(-1//x))#

Multiply the second fraction by

#= (-2e^(-1//x) + x(1 - e^(-2//x)))/(x(1 + e^(-1//x))^2)#

Let's choose

#f'(-0.1) = (-2e^(-1//-0.1) + -0.1(1 - e^(-2//-0.1)))/(-0.1(1 + e^(-1//-0.1))^2) ~~ -1#

#f'(0.1) = (-2e^(-1//0.1) + 0.1(1 - e^(-2//0.1)))/(0.1(1 + e^(-1//0.1))^2) ~~ +1#

Since the slopes very near

Therefore, this function is **NOT differentiable** at