# Need some help on differentiation

• June 14th 2006, 06:56 PM
c_323_h
Need some help on differentiation
Differentiate:

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

2. $y=\frac{t^2}{3t^2-2t+1}$

3. $y=\frac{v^3-2v\sqrt{v}}{v}$

I tried these and got 2 pages worth of arithmetic for one problem so I know I'm doing something wrong here.
• June 14th 2006, 07:27 PM
c_323_h
okay, i figured 2 out:

$\frac{t^2}{3t^2-2t+1)}=$
$\frac{(3t^2-2t+1)\frac{d}{dx}(t^2)-(t^2)\frac{d}{dx}(3t^2-2t+1)}{(3t^2-2t+1)^2}=$
$\frac{(3t^2-2t+1)(2t)-(t^2)(6t-2)}{(3t^2-2t+1)^2}=$
$\frac{2t(1-t)}{(3t^2-2t+1)^2}$
• June 14th 2006, 10:09 PM
Here is my thoughts on #1:

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

I first do some algebra with it,

Step#1 $\frac {y+5y^3}{y^2} - \frac {3y+15y^3}{y^4}$

Step#2 $\frac {y}{y^2} + \frac {5y^3}{y^2} - \frac {3y}{y^4} - \frac {15y^3}{y^4}$

Step#3 $\frac {1}{y} + 5y - \frac {3}{y^3} - \frac {15}{y}$

Step#4 $5y - 3y^{-3} - 14y^{-1}$

Now you should be able to solve it, I hope I did this right.

KK
• June 14th 2006, 10:34 PM
Problem 3:

Same thing, do the algebra first.

$
y=\frac{v^3-2v\sqrt{v}}{v}
$

Step#1 $\frac {v^3}{v} - \frac {2v\sqrt{v}}{v}$

Step#2 $v^2 - 2v^{1/2}$

Now it is just straight forward differenation.

KK
• June 15th 2006, 03:37 AM
guess
i think your partial fraction decompostion for question 2 is wrong.. please check..
• June 15th 2006, 04:00 AM
Soroban
Hello, c_323_h!

Quote:

Differentiate:

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

$2)\;y\,=\,\frac{t^2}{3t^2-2t+1}$

$3)\;y\,=\,\frac{v^3-2v\sqrt{v}}{v}$

Simplify first . . .

$1)\;\;F(y)\;=\;(y^{-2} - 3y^{-4})(y + 5y^3)\;=$ $\;y^{-1} + 5y - 3y^{-3} - 15y^{-1}\;=$ $5y - 3y^{-3} - 14y^{-1}$

$3)\;\;y\;=\;\frac{v^3 - 2v\sqrt{v}}{v} \;= \;\frac{v^3 - 2v^{\frac{3}{2}}}{v} \;=$ $v^2 - 2v^{\frac{1}{2}}$

#2 doesn't simplify, but you applied the Quotient Rule correctly . . . good work!

[By the way, pay no attention to the man behind the curtain.
Partial Fractions are used mostly for Integration, never for Differentiation.]
• June 15th 2006, 12:05 PM