Proof that f(x) = x*|x| is Differentiable at x = 0

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Actually if you graph the function u can easily see that the function is differentiable at x=0, because unlike mod function where there is a steep ascend in the negative X axis and steep ascend in the positive X axis after obtaining a value of 0 at x=0, thus giving rise to a corner point, and at corner point no function is differentiable. But the case just reverses if we take x|x|. In x|x| there is no corner corner points as when x is negative the f(x) is also negative and when x is +ve x|x| is also positive giving rise to a curve in both negative y and x axis as well as positive y and x axis.

The explanation of this video was not at all good.... Let me give you some more appropriate method. The function x|x| is definitely continuous at x=0, now do d/dx of x|x| which results in -2x when x is less than 0 and 2x when x is greater than 0. Now at x=0, both gives 0. So as both are equal hence the function is differentiable....
PS--> People confuse x|x| with |x| function. They are both different concept but look very similar. That's the beauty of mathematics.
If you have further doubt post the question on quora and request my ans there so that i can upload the image of the graph along with a solved explanation.!
*_HAPPY LEARNING_*

hirakmondal

Does the product rule fails to show x|x| to be differentiable at 0?
d/dx(x|x|) =
x * d/dx(|x|) + |x| =
2x, if x > 0 and |x| -x if x < 0 =
2|x|, if x ≠ 0

serkratos

determinant x is derivable at x=0 true or false??

aryavenugopal

At the end, you cannot just input 0 into the inequality. That's not how inequalities work. If you graph the function, you'll see that it is not differentiable at x=0

ddddtttt

Proof that f(x) = x*|x| is Differentiable at x = 0

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