Where am i going wrong on taking the derivative?

**AlephZero** · August 5th 2009, 10:56 AM

Originally Posted by skeeter

work smart, not hard.

$\frac{t^8+9t+3}{t^2} = t^6 + 9t^{-1} + 3t^{-2}$

$\frac{d}{dt}\left[t^6 + 9t^{-1} + 3t^{-2}\right] = 6t^5 - 9t^{-2} - 6t^{-3} = 6t^5 - \frac{9}{t^2} - \frac{6}{t^3}$

Originally Posted by Darkhrse99

I believe you ment $6t^5-\frac{9}{t^2}-\frac{6}{t^3}$ ?

Now, that is exactly what skeeter wrote. I cannot believe there is even still a discussion about this.

**Darkhrse99** · August 5th 2009, 10:59 AM

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