Basic concepts of Differtiation w/fractions

[Becky4paws]

I've never been confident with fraction problems and I guess these aren't going to be any different.

Directions: Differentiate the given function doing as much of the computation as possible in your head.

Problem: y=3/x - 2/x^2 + 2/3x^3

My solution (although uncomplete): y=3x^-1 - 2(2x^-1) + 2(9x^2)^-1

If my solution is right, how would I simplify that answer?

Thank you for your help, I have many more problems like this one and I hesitate to try to move forward.

 

[tkhunny]

Remember the basic polynomial rule?

(d/dx)x^a = a*x^(a-1) for a &#8800 0. Ringing any bells?

(d/dx)(3/x) = (d/dx)(3x^-1) = 3*(-1)*x^(-2) = -3/(x^2)

You do the two other pieces.

 

[soroban]

Hello, Becky4paws!

Differentiate: \(\displaystyle \,y \:= \:\frac{3}{x}\,-\,\frac{2}{x^2}\,+\,\frac{2}{3x^3}\)

My solution (although uncomplete): y=3x^-1 - 2(2x^-1) + 2(9x^2)^-1 ??

We have: \(\displaystyle \;y \;= \;3x^{-1}\,-\,2x^{-2}\,+\,\frac{2}{3}x^{-3}\)


Differentiate: \(\displaystyle \,y' \;= \;-3x^{-2}\,+\,4x^{-3}\,-\,2x^{-4}\)


Multiply "top and bottom" by \(\displaystyle x^4:\)

\(\displaystyle \L\;\;y'\;=\;\frac{x^4}{x^4}\,\cdot\,\frac{-3x^{-2}\,+\,4x^{-3}\,-\,2x^{-4}}{1} \;= \;\frac{-3x^2\,+\,4x\,-\,2}{x^4}\)