Q: What is the derivative of 2 over X?

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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 delta x and delta y (dx and dy) are the changes in x and in y. When you take the derivative you are provided with a ratio (rise over run) of these changes. Take f(x)=3x^2 + 2x Derivative: (dy/dx)=6x + 2 From this, you can solve for dx and dy algebraically.

Related questions

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.

e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.

Your expression simplifies to just x^2 {with the restriction that x > 0}. The derivative of x^2 is 2*x

X/1 is just X. so (1/2)X2 + C or X2/2 + C