Question

Find the derivative by using the product rule (negative exponents)..

Answer 1

\[(y^-1+y^-2)(2y^-3-5y^-4)\]

Answer 2

You have 2 factors. Denote
\(f(y)=y^{-1}+y^{-2}\\
g(y)=2y^{-3}-5y^{-4}\)

Then the product rule states:
\((f(y)\cdot g(y))' = f'(y)g(y) + f(y)g'(y)\)

Answer 3

And you can find the derivatives \(f'(y)\) and \(g'(y)\) easily by using the power rule.
\[ \large \frac{d}{dy}y^n=ny^{n-1}\]

Answer 4

So my product would be:
\[-1y^-2y^-3(2y^-3-5y^-4)+ 6y^-4+20y^-5(y^-1+y^-2)\]
and I would just distribute ??

Answer 5

Looks almost good, not sure if the first part is a typo or not, but should be \(-1y^{-2}-2y^{-3}\) Also. Just just make sure you keep each element of the product rule in parentheses. I.e. write \((-1y^{-2}-2y^{-3})(2y^{-3}-5y^{-4})\) and same on the other side of the +

Answer 6

also after the plus, it should be \(-6y^{-4}\) since
\[\frac{d}{dy}2y^{-3}=2[(-3)y^{-4}]=-6y^{-4}\]

Answer 7

Yes, I think I made a few typos. I'm double checking right now...

Answer 8

Ok, so I got:
\[-8y^-5+15y^-6+30y^7\]
as my final answer (:

Answer 9

I think you mean \(30y^{-7}\) :)

Answer 10

otherwise it's good

Answer 11

Oh just a tip for your LaTeX-ing, if you want to keep the exponents in the right format, write:
y^{-4} rather than just y^-4, and it will put the - and 4 in the exponent

Answer 12

Ok, thank you very much for all the help (: