I know this is very simple but I want to make sure I did it right. The derivative of f(x)= 1/(x^2) is 1 or 1/2?

Question

datanewb · Answer

The equation:
$$f(x) = \frac{1}{x^2} $$can also be written$$f(x) = x^{-2}$$. Taking the derivative of this (the exponent will be -2-1 = -3, and then we multiply by the original exponent to get:
$$f'(x) = -2x^{-3}$$ Showing this again as a fraction would look like$$f'(x) = \frac{-2}{x^3}$$

Note you could have also used the division rule $$\left(\frac{u}{v}\right)' = \frac{u'v-v'u}{v^2} $$
This would look like $$f'(x) =\frac{0*x^2-1*2x}{x^4}\=\frac{-2x}{x^4}\=\frac{-2}{x^3}$$

anonymous · Answer

I see, so always flip the fraction when the exponent is in the denominator to work out the derivative?

datanewb · Answer

Well, you don't have to flip them (for instance you can use the quotient rule for derivatives instead), but I often find it easier to rewrite the equation with the variable in the exponent.  Then it is really a simple straight forward process.  

Thanks for the medal.