1. ## nondifferentiable?

Is the graph of $\displaystyle y=\frac{1}{x}$ nondifferentiable?

2. Originally Posted by live_laugh_luv27
Is the graph of $\displaystyle y=\frac{1}{x}$ nondifferentiable?
First, note that you have $\displaystyle y = x^{-1}$. you can use the power rule to find the derivative where it exists.

to find where it does not exist, recall that a function $\displaystyle f(x)$ is differentiable at all $\displaystyle x$ where the limit

$\displaystyle f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}h$

exists

3. Originally Posted by live_laugh_luv27
Is the graph of $\displaystyle y=\frac{1}{x}$ nondifferentiable?
the function, y = 1/x , is not differentiable at x = 0 because it is not defined there.

4. ok..
I'm looking at a graph, and it asks how many values are nondifferentiable. One is clearly a corner, and one is clearly a discontinuity. But the third function is just a curve, similar to $\displaystyle y=\frac{1}{x}$, and I can't tell if it is a vertical tangent or not.

5. Originally Posted by live_laugh_luv27
ok..
I'm looking at a graph, and it asks how many values are nondifferentiable. One is clearly a corner, and one is clearly a discontinuity. But the third function is just a curve, similar to $\displaystyle y=\frac{1}{x}$, and I can't tell if it is a vertical tangent or not.
there is no "corner" on the graph of 1/x. it is undefined (and hence, not differentiable) at x = 0. this we can tell just by looking at the formula for the function.

in general though, you would examine the limit i gave you. note that we have here:

$\displaystyle f'(x) = \lim_{h \to 0} \frac {\frac 1{x + h} - \frac 1x}h = \lim_{h \to 0} \frac {-h}{h(x(x + h))} = - \frac 1{x^2}$

clearly if $\displaystyle x = 0$ the limit is undefined, otherwise, it is good though