The expression ( X^\pi ) is undefined for negative values of ( X ) when ( \pi ) is not an integer because it involves taking a root of a negative number, which can lead to complex results. For non-integer exponents, the operation requires a real base, and negative bases with non-integer exponents cannot be simplified to real numbers. Specifically, the result would be a complex number, making the expression undefined in the context of real numbers.
In the domain [0, 2*pi],sin is negative for pi < x < 2*picos is negative pi/2 < x < 3*pi/2 andtan is negative for pi/2 < x < pi and 3*pi/2 < x < pi.Also, the same applies for all intervals obtained by adding any integer multiple of 2*pi to the bounds.
x(pi+1)/(pi+1)
The logarithm of a negative number is undefined in the realm of real numbers. This is because logarithmic functions are defined only for positive real numbers. However, in the context of complex numbers, the logarithm can be defined for negative numbers using the formula ( \log(-x) = \log(x) + i\pi ), where ( i ) is the imaginary unit. This allows for a complex representation of logarithms of negative values.
3
since x is negative you use the identity cot-1(x)=tan-1(1/x)+pi. Tan-1(1/-sqrt3) + pi 5pi/6 + pi =pi
The negative reciprocal of zero would be any undefined fraction. 0/x=0 negative reciprocal: -(x/0)= undefined.
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.
In the domain [0, 2*pi],sin is negative for pi < x < 2*picos is negative pi/2 < x < 3*pi/2 andtan is negative for pi/2 < x < pi and 3*pi/2 < x < pi.Also, the same applies for all intervals obtained by adding any integer multiple of 2*pi to the bounds.
x(pi+1)/(pi+1)
3
For 0 < x < pi. sin(x) is positive,for pi < x < 2*pi, sin(x) is negative and these intervals can be left or right-shifted by any multiple of 2*pi radians.
since x is negative you use the identity cot-1(x)=tan-1(1/x)+pi. Tan-1(1/-sqrt3) + pi 5pi/6 + pi =pi
The Yth power of X, or XY is the value you get when you multiply Y number of X's together. If X is equal to 1, however, it does not matter how many 1's you multiply - the answer is alway 1, therefore 1X is not useful and is undefined.
mass of proton is 6 x pi raised to power 5 times mass of electron
y=x^pid/dx=pi*(x^pi-1)This is true because of power rule.d/dx (x^a)=a(x^(a-1))
-8. alpha x 17.81
A negative power. When something is to the power of negative two, say X it would be 1/X. If something is to the power of negative 3, it would be.... etc