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f(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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9y ago
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15y ago

negative one-half.

If you understand the meaning of derivative, you will realize that the question is the same as asking "What is the slope of the line y = - 1/2 x"

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Q: What is the derivative of 2 over X?
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