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I have no idea what I have done wrong. Please criticise. enter image description here

graydad
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Cetshwayo
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1 Answers1

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You are just off by a negative sign. You say that you say $\frac{d}{dx} [6-\cot(x)]= -\csc^2(x)$, but it should be $\frac{d}{dx} [6-\cot(x)] = \csc^2(x)$. This means you should get $$y' = \frac{42-7\cot(x)-7x\csc^2(x)}{(6-\cot(x))^2}$$ Does this match what you would expect?

graydad
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  • I shouldn't distribute the negative sign? – Cetshwayo Mar 13 '15 at 20:00
  • Well, there isn't really one to distribute since you have $(-7x) \cdot (\csc^2(x))$ – graydad Mar 13 '15 at 20:01
  • Okay, the second line of my work, after I calculated the derivative of 6-cot x; which is 0 - (-csc^2x), is there no need to distribute the negative sign? May be this is a basic math problem; however, I would like some to explain to me if I am doing "illegal math". – Cetshwayo Mar 13 '15 at 20:07
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    Nothing illegal going on, $-(-\csc^2(x)) = \csc^2(x)$. Then you still have to multiply that quantity by $-7x$, so the negative sign appears again. $$-7x(0-(-\csc^2(x))) = -7x\cdot 0 -7x \cdot -(-\csc^2(x)) = 0-7x(\csc^2(x)) = -7x\csc^2(x)$$ – graydad Mar 13 '15 at 20:08
  • So -(a)(b) not -[(a)(b)] – Cetshwayo Mar 13 '15 at 20:11
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    There is no difference between those two – graydad Mar 13 '15 at 20:12