Problem - find second derivative

1. Dec 15, 2012

jkristia

The problem and my attempt at a solution is shown in the attached image.

The problem is that I end up with one extra x in the denominator.
So the question is: Is my expression for y'' correct and I just made a mistake somewhere (I have checked it several times), or am I missing something in the expression for y' and/or y'' ?

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2. Dec 16, 2012

MarneMath

If the answer in the book is something like 6/x^3 + 18/x^4 + 12/x^5, you got the same answer.

3. Dec 16, 2012

SammyS

Staff Emeritus
How do you go from $\displaystyle\ \frac{6}{x^3}\left( 1+\frac{3}{x}+\frac{2}{x^2}\right)$/extract_itex] to $\displaystyle\ \frac{6}{x^4}\left( 1+\frac{1}{x}\right)\left( 1+\frac{2}{x}\right)\ ?$ $\displaystyle\ 1+\frac{3}{x}+\frac{2}{x^2}=\left( 1+\frac{1}{x}\right)\left( 1+\frac{2}{x}\right)\ .$ 4. Dec 16, 2012 symbolipoint Does this make sense? $$\[ \begin{array}{l} y' = 3(1 + {\textstyle{1 \over x}})^2 x^{ - 2} ( - 1) \\ = \frac{{ - 3(1 + {\textstyle{1 \over x}})^2 }}{{x^2 }} \\ \end{array}$$$

Next you would use quotient rule to find the next derivative (second derivative).

5. Dec 16, 2012

jkristia

6. Dec 16, 2012

Curious3141

Personally, I'd just expand the binomial then differentiate twice. Keep the exponents as negative powers, it makes everything quicker.

7. Dec 16, 2012

jkristia

>>Personally, I'd just expand the binomial then differentiate twice
That certainly makes it much easier. I guess I didnt think of that since I was doing problems related to the chain rule. Thank you for your suggestion.