1. ## Derivative help?.. again

$F(y)=(\frac{1}{y^2}-\frac{3}{y^4})(y+5y^3)$

I'm not sure where to start so I tried to get rid of the fraction.
$
(y^4)(\frac{1}{y^2}-\frac{3}{y^4})(y+5y^3)$

$=(y^2-3)(y^5+5y^7)$
now I differentiated
$(2y-0)(y^5+5y^7)+(y^2-3)(5y^4+35y^6)$
which turns into a ridiculously large.. and wrong answer after i simplify $
45y^8-38y^6-15y^4$

book's answer is $
5+\frac{14}{y^2}+\frac{9}{y^4}$

2. you also must divide by $y^4$
the derivative of $f(y)$ is not the same as $y^4f(y)$
Just multiply the original function and differentiate term by term, and don't use the product rule.

3. Originally Posted by dorkymichelle
$F(y)=(\frac{1}{y^2}-\frac{3}{y^4})(y+5y^3)$

I'm not sure where to start so I tried to get rid of the fraction.
$
(y^4)(\frac{1}{y^2}-\frac{3}{y^4})(y+5y^3)$

$=(y^2-3)(y^5+5y^7)$
now I differentiated
$(2y-0)(y^5+5y^7)+(y^2-3)(5y^4+35y^6)$
which turns into a ridiculously large.. and wrong answer after i simplify $
45y^8-38y^6-15y^4$

book's answer is $
5+\frac{14}{y^2}+\frac{9}{y^4}$
on way to get rid of fractions is to inverse the exponents
$f(y) = (y^{-2}-3y^{-4})(y+5y^{3})$
then FOIL it
f $(y) = y^{-1}+5y-3y^{-3}-15y^{-1}
\Rightarrow
5y-14y^{-1}-3y^{-3}$

so then $f'(y) = 5+14y^{-2}+9y^{-4}\Rightarrow5+\frac{14}{y^2}+\frac{9}{y^4}$

4. So I don't use the multiplication derivative rule?

5. ## rewrite expression into terms

i would agree with matheagle that you can use the product or quotient rule
but it is much more efficient to rewrite your expression into just terms and take the derivative from there.

$(f \cdot g)' = f \cdot g'+g \cdot f'$

and

$\left(\frac{f}{g}\right)' = \frac{g \cdot f'-f \cdot g'}{g^2}$

would be quite involved with this problem

hope this helps.

6. I would just use the power rule as wavelet did