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!
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.
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.
Your expression simplifies to just x^2 {with the restriction that x > 0}. The derivative of x^2 is 2*x
The derivative is 2x based on the power rule. Multiply the power by the coefficient of x then drop the power by one.
the derivative of ln x = x'/x; the derivative of 1 is 0 so the answer is 500(1/x)+0 = 500/x
The first derivative of e to the x power is e to the power of x.
The derivative of x is not x. X is the same as x^1, so you use the power rule which decreases the power by one and brings the exponent down, giving 1x^0, which is equal to 1.
f(x)=1 f'(x)=0 because the derivative of a constant is ALWAYS 0.
The derivative of ex is ex
Regardless of what 'x' is, (x)0 = 1 . tan(1 radian) = 1.55741 (rounded) tan(1 degree) = 0.01745 (rounded) We can't remember the derivative of the tangent right now, but it doesn't matter. This particular tangent is a constant, so its derivative is zero.
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.
The absolute value of x, |x|, is defined as |x| = x, x>=0; -x, x<0. If you derive this, then you will find that the derivative is 1 when x>=0, and -1 when x<0. But this means that the derivative as x approaches 0 from the left does not equal the derivative as x approaches 0 from the right, as -1=/=1. So the limit as x approaches 0 does not exist, and therefore the gradient does not exist at that point, and so |x| cannot be differentiated at x = 0.
2x is the first derivative of x2.
e^(-2x) * -2 The derivative of e^F(x) is e^F(x) times the derivative of F(x)