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How would I figure out the following problem.


$\frac{d^2}{dx^2}$ $[(x^3-1)\frac{d^2}{dx^2}(6x-x^5)$

This is what I did

I took the derivative of $(6x-x^5)$ and got $(-20x^3)$

I did $(x^3-1)(-20x^3)$ and got $(-20x^6+20x^3)$

Then I took the derivative of that twice and got


But did I do this correctly?

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Yes you did. (But, the first thing you did was to take the second derivative of $6x-x^5$, right?) – David Mitra Feb 9 '13 at 23:26
Yes indeed. That is what I did. – Fernando Martinez Feb 9 '13 at 23:29
where does the square bracket close ] ? – Elements in Space Feb 9 '13 at 23:42
up vote 3 down vote accepted

Assuming you mean to close the left bracket on the right of the entire expression:$$\frac{d^2}{dx^2}\left[(x^3-1)\frac{d^2}{dx^2}(6x-x^5)\right]$$

Then your result is correct, but I suspect you meant that you took the derivative twice (sequentially), of $(6x - x^5)$, getting $$d/dx (6 - 5x^4) = -20x^3.$$

Then you correctly multiplied $(x^3 - 1)(-20x^3)$.

And then took the derivative of that product, twice, to get your answer: $-600x^4 + 120x$

Nice job.

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I wish, Leibniz had been here, to see his symbol has been used so widely these days. I think he would give you a plus. – Babak S. Feb 10 '13 at 3:43

Take the derivative of the derivative of $6x-x^5$, then multiply the result by $x^3-1$, and then take the derivative of the derivative of that.

You should get $120x-600x^4$.

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yes that is what I got. I think its correct. – Fernando Martinez Feb 9 '13 at 23:31

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