f(x)=(pi2)x=pi2x. The derivative of kx=ln(k)*kx, so f'(x)=2ln(pi)*pi2x (with chain rule).
The derivative with respect to 'x' of sin(pi x) ispi cos(pi x)
pi cos(pi x)
The anti-derivative of any constant c, is just c*x. Thus, the antiderivative of pi is pi*x. We can verify this by taking the derivative of pi*x, which gives us pi.
The first derivative of e to the x power is e to the power of x.
x(pi+1)/(pi+1)
-(pi)*sin(pi*x)
y = Sin(pi) = 0 Then its derivative is dy/dx = Cos(pi). = -1
pi divided by 6 is a constant and so its first derivative is 0. And since that is also a constant, the second derivative is 0. It is not clear what f(x) = csc(x) has to do with that!
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.
mass of proton is 6 x pi raised to power 5 times mass of electron
The first deriviative of sin(x) is cos(x), which is also sin(x + pi/2). The general formula, then for the nth deriviative of sin(x) is sin(x + n pi/2).