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I'm doing a problem in my textbook that I have the answer too thanks to it being an odd problem in the back. The problem is that I can't get to the answer of this problem.

Problem by the way is shown to be "Differentiate: (1 / y^2 - 3 / y^4)(y + 5y^3)".

The answer in the book happens to be: "5 + 14/ y^2 + 9/Y^4".

What I've tried thus far is use (fg)' = fg' + gf'

With F being the first parenthesis and G being the second parentheses respectably.

But doing so gets me (y^2 - 3)(5y^2 + 1) / y^3 + (y^2 - 3)(5y^2 + 1). I stop at this point because I know theirs no way to get the answer from here.

Can someone point me to the right way?Also how can I make fractions like others are doing on this site? I can't find the button on here..

$\frac{1}{y^2}$

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    Try multiplying first then find the derivative – 2017-02-12
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    The quotient rule is really helpful for quotients of non constant functions. When one has a constant in the numerator, one can simply use the constant rule. – 2017-02-12
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    `how can I make fractions like others are doing` Lookup the [MathJax](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) basics. For example, type `$\frac{1}{y^2}$` to get $\frac{1}{y^2}$. – 2017-02-12
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    @kingW3 So far looks promising I think I can use the derivative. I'm at a point 5y^4 + 14y^2 + 9 all over y^4. It's very close but the answer dictates something a little different. Can I re-write it in someway? – 2017-02-12
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    @JoeJack3man75 Please check again. When applying $\frac{d}{dy} (y^n)=ny^{n-1}$, you should be getting negative powers of $y$. – 2017-02-12

2 Answers 2

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You can expand what you have to give: $$\left(\frac{1}{y^2}-\frac{3}{y^4}\right)\left(y+5y^3\right)=5y-14y^{-1}-3y^{-3}$$ Therefore, the derivative will be given by: $$\frac{d}{dy} (5y-14y^{-1}-3y^{-3})=5\cdot\frac{d}{dy}(y)-14\cdot\frac{d}{dy}(y^{-1})-3\cdot\frac{d}{dy}(y^{-3})$$ Now, use the fact that $\frac{d}{dy}(y^n)=ny^{n-1}$.

Alternatively, you can use the product rule or the quotient rule, as you have mentioned. I personally prefer to expand it in this case.

I see that you asked how to format questions using LaTeX. Here is a nice guide showing how to use the formatting.

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    Oh god the skinning of multiple ways! Lol jk. Can you show more steps pease? – 2017-02-12
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    @JoeJack3man75 The answer is just one step away. Try differentiating $5y$ with the rule I've provided. In this case, set $n=1$. Differentiate the other terms. – 2017-02-12
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    Just tried now and nothing....It is /over that helps create the fraction right? I've tried /frac 1Y^2 also but no dice. – 2017-02-12
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    @JoeJack3man75 Use the backward slash "\" instead "\frac{1}{y^2}" – 2017-02-12
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    Anytime you can write a product or a quotient as the sum or difference of powers, you can eliminate the use of the product rule or the quotient rule for simply the sum and difference rules combined with a power rule. – 2017-02-12
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So, you are taking the derivative with respect to y. What you have to do is to first expand your expression by multiplying it out. You will get

$\frac{1+5y^2}{y}-\frac{3+15y^2}{y^3}$ after some simplification.

Then you simply take the derivative using the quotient rule: $\frac{f'g-fg'}{g^2}$ and you'll have it!

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    ((1+5(y^2))/y)-((3+15(y^2))/(y^3)) Can you make this clearer? I'm having trouble seeing this on paper – 2017-02-12