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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.
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Taigaf(x) = 2/x = 2x^-1
so dy/dx = -2x^-2 = -2/x²
f(x) = 2/x
This can be re-written as f(x) = 2x^-1 according the to the laws of exponents. Now, we just take the derivative normally:
f(x) = 2x^-1
f ' (x) = -1 × 2x^(-1-1) = -2x^-2
The above can be re-written in quotient form as f ' (x) = -2/x²
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What is the derivative of 1 over x?
Derivative of 1/x 1/x = x-1 Take the derivative (-1)x(-1-1) = -x-2 = 1/x2
What is the derivative of 2 to the power of x?
if f(x)=kx, f'(x)=ln(k)*kx. Therefore, the derivative of 2x is ln(2)*2x.
What is the derivative of 1 divided by root x?
(1/2(x^-1/2))/x
How do you find the second derivative?
Afetr you take the first derivative you take it again Example y = x^2 dy/dx = 2x ( first derivative) d2y/dx2 = 2 ( second derivative)
What is the anti derivative of x squared plus x?
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
