The derivative of 2/x can be found using the quotient rule in calculus. The quotient rule states that the derivative of f(x)/g(x) is [g(x)f'(x) - f(x)g'(x)] / [g(x)]^2. Applying this rule to 2/x, where f(x) = 2 and g(x) = x, the derivative is calculated as [x0 - 21] / x^2, which simplifies to -2/x^2. Therefore, the derivative of 2/x is -2/x^2.
Derivative of 1/x 1/x = x-1 Take the derivative (-1)x(-1-1) = -x-2 = 1/x2
if f(x)=kx, f'(x)=ln(k)*kx. Therefore, the derivative of 2x is ln(2)*2x.
(1/2(x^-1/2))/x
Afetr you take the first derivative you take it again Example y = x^2 dy/dx = 2x ( first derivative) d2y/dx2 = 2 ( second derivative)
The anti-derivative of X2 plus X is the same as the anti-derivative of X2 plus the anti-derivative of X. The anti derivative of X2 is X3/3 plus an integration constant C1 The anti derivative of X is X2/2 plus an integration constant C2 So the anti-derivative of X2+X is (X3/3)+(X2/2)+C1+C2 The constants can be combined and the fraction can combined by using a common denominator leaving (2X3/6)+(3X2/6)+C X2/6 can be factored out leaving (X2/6)(2X+3)+C Hope that helps
Derivative of 1/x 1/x = x-1 Take the derivative (-1)x(-1-1) = -x-2 = 1/x2
The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x
Derivative of x = 1, and since sqrt(x) = x^(1/2), derivative of x^(1/2) = (1/2)*(x^(-1/2))Add these two terms together and derivative = 1 + 1/(2*sqrt(x))
-2
x*x1/2= x3/2 Derivative = 3/2 * x1/2
1/x = x-1d/dx(x-1) = -x-2 = -1/x2
-1/2*x-3/2 which is equal to -1/[2*x3/2]
e^(-2x) * -2 The derivative of e^F(x) is e^F(x) times the derivative of F(x)
2 x 2 = 4. 4 is a constant. The derivative of a constant is always 0. Therefore, The derivative of 2 x 2 is zero.
Your expression simplifies to just x^2 {with the restriction that x > 0}. The derivative of x^2 is 2*x
e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.
X/1 is just X. so (1/2)X2 + C or X2/2 + C