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y=x^pi

d/dx=pi*(x^pi-1)

This is true because of power rule.

d/dx (x^a)=a(x^(a-1))

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derivative is: ln(pi)pix dx

Q: What is the derivative of pi raised to x?

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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)

Sine (pi) is a constant, so the derivative of sine (pi) is zero.

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).