I need to find the Derivative of $\displaystyle f(x)= x+1/x^2$ Can I do this? $\displaystyle x=1/x^2= x-x^2=-x^2+x=f'(x)= -2x+1$ Thanks Jason
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Originally Posted by Darkhrse99 I need to find the Derivative of $\displaystyle f(x)= x+1/x^2$ Thanks Jason $\displaystyle f(x) = x + x^{-2} $ $\displaystyle f'(x)=1 +(-2)x^{-2-1}=1-2x^{-3}$
Originally Posted by Darkhrse99 I need to find the Derivative of $\displaystyle f(x)= x+1/x^2$ Thanks Jason You can write this as $\displaystyle f(x)= x + x^{-2}$ Using your normal differential rules this gives you $\displaystyle f'(x)= 1 - 2x^{-3}$
Originally Posted by Amer $\displaystyle f(x) = x + x^{-2} $ $\displaystyle f'(x)=1 +(-2)x^{-2-1}=1-2x^{-3}$ Just beat me to it
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