# Slope of tangent at…

Can anyone check my math on the following equation? I'm trying to find the slope of the tangent at$x=3.5$ for $$y = 0.000005x^4 - 0.0014x^3 + 0.1007x^2 - 4.4776x + 168.79$$

It's been a very long time since I've done calculus, and I THINK what I need to do is find the derivative before plugging$x=3.5$ in...

$$y' = 0.00002x^3 - 0.0042x^2 + 0.2014x - 4.4776$$ $$y'(3.5) = 0.000000000000343 - 0.00021609 + 0.7049 - 4.4776$$ $$y'(3.5) = -3.772916089999657$$

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No problem with derivative. Problem with substitution. For example first term should be $0.0008575$. – André Nicolas Oct 9 '12 at 16:33
as long as you did not make a numerical mistake, it should be correct... – David Hoffman Oct 9 '12 at 16:41
After corrections, I think it is this... y'(3.5) = -4.142085625 – Szuturon Oct 9 '12 at 18:36
Your previous answer was closer to the truth. I get about $-3.8232925$. No guarantees! – André Nicolas Oct 9 '12 at 18:48
Ok. I don't know why I didn't just do this from the start. I went ahead and plugged in "=0.00002*(3.5^3)-0.0042*(3.5^2)+0.2014*(3.5)-4.4776" into Excel. I have -3.8232925 now. I suck. – Szuturon Oct 9 '12 at 20:01

Your derivative was calculated correctly. The substitution has mistakes, due to incorrect use of the calculator. For example, you wanted to evaluate $0.00002x^3$ where $x=3.5$. What you did is almost certainly to multiply $0.00002$ by $3.5$, and then take the power. So you calculated $(0.00002x)^3$. This is $(0.00002)^3(3.5)^3$, very tiny.
What you need to do is to calculate $(3.5)^3$ first, and then multiply the result by $0.00002$. There is a similar issue with the evaluation of the next term.
@Szuturon: It was just a minor slip of the fingers, much more likely in this calculator age than it used to be. About the $1/3$, I would probably give full marks, unless the slip results in an answer to an "applied" problem that is physically unreasonable. And I would still give full marks if the person said it is unreasonable. But with a multiple choice or "exact answer" test, so undeservedly popular nowadays, the mark would probably be $0$. Another reason not to have such tests. – André Nicolas Oct 9 '12 at 18:38