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    Post nondifferentiable?

    Is the graph of y=\frac{1}{x} nondifferentiable?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by live_laugh_luv27 View Post
    Is the graph of y=\frac{1}{x} nondifferentiable?
    First, note that you have 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 f(x) is differentiable at all x where the limit

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

    exists
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    Quote Originally Posted by live_laugh_luv27 View Post
    Is the graph of y=\frac{1}{x} nondifferentiable?
    the function, y = 1/x , is not differentiable at x = 0 because it is not defined there.
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    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 y=\frac{1}{x}, and I can't tell if it is a vertical tangent or not.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by live_laugh_luv27 View Post
    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 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:

    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 x = 0 the limit is undefined, otherwise, it is good though
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