y"+y'=0 is a differential equation and mean the first derivative plus the second derivative =0.Look at e-x the first derivative is -e-xThe second derivative will be e-xThe sum will be 0
the derivative of ln x = x'/x; the derivative of 1 is 0 so the answer is 500(1/x)+0 = 500/x
The derivative if a function is basically it's slope, or its rate of change. An example is the function y = 4x - 6. This is a line with a slope of 4. The derivative is y' = 4. Another example is the function y = 3x2. This is a parabola with a vertex at (0,0). Its derivative is y' = 6x. At x = 0, the slope of the parabola is 6*0, which is 0, since this is the vertex of the parabola. To the left, at x is -4 for example, the derivative (and therefore slope) is negative. To the right, at x = 5 for example, the derivative is positive. The farther away from the vertex, the greater the value of the derivative so the the slope of the function increases as you move away from the vertex (it gets steeper).
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 derivate of zero - as well as the derivative of ANY constant (non-variable) number, is zero. (A graph of y = 0 for example will be a horizontal line - the slope is zero.)
A dot A = A2 do a derivative of both sides derivative (A) dot A + A dot derivative(A) =0 2(derivative (A) dot A)=0 (derivative (A) dot A)=0 A * derivative (A) * cos (theta) =0 => theta =90 A and derivative (A) are perpendicular
the derivative is 0. the derivative of a constant is always 0.
The derivate of 3x is 3; the derivative of -1 is 0. So, the derivative of 3x-1 is simply 3.The derivate of 3x is 3; the derivative of -1 is 0. So, the derivative of 3x-1 is simply 3.The derivate of 3x is 3; the derivative of -1 is 0. So, the derivative of 3x-1 is simply 3.The derivate of 3x is 3; the derivative of -1 is 0. So, the derivative of 3x-1 is simply 3.
The derivative of all constants is 0.
0 A derivative is the rate of change of a function as another variable changes. As there is no change to a constant, the derivative is necessarily 0.
ee is a constant and so its derivative is 0.
y"+y'=0 is a differential equation and mean the first derivative plus the second derivative =0.Look at e-x the first derivative is -e-xThe second derivative will be e-xThe sum will be 0
x^0 = 1 for all x. The derivative of 1 is always zero.
the derivative of ln x = x'/x; the derivative of 1 is 0 so the answer is 500(1/x)+0 = 500/x
derivative of a constant is 0, because the function of a constant is a line with no change in slope, so d2011/dx=0
f(x)=1 f'(x)=0 because the derivative of a constant is ALWAYS 0.
The first derivative is m and the second is 0 so the third is also 0.