# 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.