x^0 = 1 for all x.
The derivative of 1 is always zero.
Chat with our AI personalities
Your expression simplifies to just x^2 {with the restriction that x > 0}. The derivative of x^2 is 2*x
e^(-2x) * -2 The derivative of e^F(x) is e^F(x) times the derivative of F(x)
The derivative of a constant is always 0. To show this, let's apply the definition of derivative. Recall that the definition of derivative is: f'(x) = lim h→0 (f(x + h) - f(x))/h Let f(x) = 1. Then: f'(x) = lim h→0 (1 - 1)/h = lim h→0 0/h = lim h→0 0 = 0!
the derivative of 1x would be 1 the derivative of x to the power of 1 would be 1. the derivative of x+1 would be 1 the derivative of x-1 would be 1 im not sure what you are asking, but however you put it, it's 1.
When you take the derivative of a function, you are seeking a variation of that function that provides you with the slope of the tangent (instantaneous slope) at any value of (x). For example, the derivative of the function f(x)=x^2 is f'(x)=2x. Notice that the derivative is denoted by the apostrophe inside the f and (x). Also note that at x=0, f'(x)=0, which means that at x=0 the slope of the tangent is zero, which is correct for the function y=x^2.