# Is the function differentiable at x = p?

1. Jan 8, 2016

### PcumP_Ravenclaw

Hello mates,

is the function $f(x) = \frac{(2^x - 1)}{x}$ differentiable at x = 0? For it to be differentiable it has to be continuous? From the graph f(0) is undefined although limit exists. I have read that at points like a corner, gap and vertical tangents it is not differentiable. So why is it differentiable at f(0)?

Sehr Danke!

2. Jan 8, 2016

### Isaac0427

Quotient rule:
if $f(x) = g(x) / h(x)$, $g(x) = 2^x - 1$ and $h(x) = x$ then $f'(x) = (h(x) * g'(x) - g(x) * h'(x) / g(x)^2$. In this case, $g(x)^2 = 0$ so the derivative would be undefined. Does this help?

3. Jan 8, 2016

### Staff: Mentor

No.
This function isn't defined at x = 0, hence it's not continuous at x = 0, so it can't be differentiable there.

4. Jan 8, 2016

### SammyS

Staff Emeritus
However, $\displaystyle \ \lim _{x\to0} f'(x)\$ does exist.

5. Jan 8, 2016

### WWGD

This assumes $g \neq 0$ which is not the case, like many said. You may say that _at points where $g \neq 0$_ (since both $f,g$ are everywhere differentiable), you can say $(f/g)' = \frac {f'g-fg'}{g^2}= \frac {(ln2)(2^x)x -2^x}{x^2}$, using the identity $2^x = e^{xln2}$.

6. Jan 9, 2016

### pasmith

Functions can only be differentiable where they are actually defined. However the singularity at the origin is removable by setting $f(0) = \ln 2$. Then the function is not merely differentiable at the origin but analytic, and its Taylor series is $$f(x) = \sum_{n=0}^\infty \frac{(\ln 2)^{n+1} x^n}{(n+1)!}.$$