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x^0 = 1 for all x.

The derivative of 1 is always zero.

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Q: What is the derivative for x to the power of 0?
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Your expression simplifies to just x^2 {with the restriction that x > 0}. The derivative of x^2 is 2*x


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What is the derivative of sin x power 0?

I think you are asking "what is the derivative of [sin(x)]^0=sin^0(x)?" and I shall answer this accordingly. Recall that x^0 = 1 whenever x is not 0. On the other hand, also notice that 0^0 is generally left undefined. Thus, sin^0(x) is the function f(x) such that f(x) is undefined when x = n(pi) and 1 everywhere else. As a result, on every open interval not containing a multiple of pi, i.e. on (n(pi), (n+1)(pi)) the derivative will be zero, since f is just a constant function on these intervals, and whenever x is a multiple of pi, the derivative at x will be undefined. Thus, [d/dx]sin^0(x) is undefined whenever x = n(pi) and 0 everywhere else. In some cases, mathematicians define 0^0 to be 1, and if we were to use this convention, sin^0(x) = 1 for all x, and its derivative would just be 0.


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Why absolute value of x is not differentiable at point 0?

The absolute value of x, |x|, is defined as |x| = x, x>=0; -x, x<0. If you derive this, then you will find that the derivative is 1 when x>=0, and -1 when x<0. But this means that the derivative as x approaches 0 from the left does not equal the derivative as x approaches 0 from the right, as -1=/=1. So the limit as x approaches 0 does not exist, and therefore the gradient does not exist at that point, and so |x| cannot be differentiated at x = 0.


What is the derivative of e to the power of -2 times x?

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