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How would I figure out the following question.

Find $f''(x)$ if $f(x)=(x^2-6x-7)^{11}$

Using the chain rule I got the first derivative as.

$11(x^2-6x-7)^{10}(2x-6)$

Applying both the chain rule and the product rule I got

for my second derivative

$f''(x)=11(x^2-6x-7)^{10}(2)+(2x-6)(110)(x^2-6x-7)^9(2x-6)$

However did I do this correctly?

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    $\begingroup$ It looks good to me. Typo, $f(x)''$ in last row. $\endgroup$ – Cortizol Feb 15 '13 at 18:42
  • $\begingroup$ The "prime" should come after the $f$. It is the function that you are differentiating. Hence $f'(x)$ and $f''(x)$ are correct while $f(x)'$ and $f(x)''$ are incorrect. $\endgroup$ – Fly by Night Feb 15 '13 at 18:47
  • $\begingroup$ yes that is true. $\endgroup$ – Fernando Martinez Feb 15 '13 at 18:48
  • $\begingroup$ Yes! $\endgroup$ – P.. Feb 15 '13 at 18:50
  • $\begingroup$ Hmm in wolfram alpha it is written differently but I guess it is simplified If I try and simplify it I will probably make an error... $\endgroup$ – Fernando Martinez Feb 15 '13 at 18:54
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Yes, your answer is correct.

Alternatively you could write $x^2-6x-7=(x+1)(x-7)$ so you can differentiate twice: $$(x+1)^{11}(x-7)^{11}$$

using just the product rule.

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You can further simplify it as

$(-x^2 + 6x + 7)^9(-22x^2 + 132x - 440(x - 3)^2 + 154)$

or rearranging signs

$(x^2 - 6x - 7)^9(22x^2 - 132x + 440(x - 3)^2 - 154)$

and finally

$22(x^2 - 6x - 7)^9(21x^2 - 126x + 173)$

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  • $\begingroup$ Your simplify routine seems to add a few too many minus signs. $\endgroup$ – Thomas Andrews Feb 15 '13 at 19:03

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