Is the function differentiable at x = p?

[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!

 

[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?

 

[Mark44]

Hello mates,

is the function $f(x) = \frac{(2^x - 1)}{x}$ differentiable at x = 0?

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

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!

 

[SammyS]

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

 

[WWGD]

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?

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 $2^x = e^{xln2}$.

 

[pasmith]

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!

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 $$f(x) = \sum_{n=0}^\infty \frac{(\ln 2)^{n+1} x^n}{(n+1)!}.$$