Prove, using the product rule that, the derivative of x^{n} is nx^{n-1} where n is a natural number. What if n is an integer or n is rational?

We know that d / dx (x) = 1. Looking at x^2 as a specific example, and using the product rule, we see that d/dx(x^2) = xd/dx(x) + d/dx(x)x = x + x = 2x. Similarly for x^3, d/dx(x^3) = 3 * x^2 * d/dx(x) = 3 x^2 Following this logic for general n, we have that d/dx(x^n) = nx^(n-1).If n is negative then we need to use that the derivative of 1 / x = - 1 / x^2. To show this, let y = 1/x so xy = 1. Then differentiate with respect to x to get that xdy/dx + y = 0, and so xdy/dx = -1 / x which is what we need.We can now proceed in the same way. d/dx(1 / x^-n) = -n * 1 / x^(-n - 1) * -1 / x^2 = nx^(n-1).For rational n we need a slightly different approach. Let y = x^n = x^(a/b) where a and b are integers, so we have that y^b = x^a. Differentiating this with respect to x gives, using the previous parts, that by^(b-1)dy/dx = ax^(a-1) and so dy/dx = (a/b)(x^(a - 1) / y^(b-1)) = (a / b)(x^(a-1) / x^(a(b-1) / b)) = (a/b)x^((a - 1) - a(b-1)/b) = (a/b)(x^(ba - b - ab + a) / b) = (a/b)x^(a / b - 1).