Derivatives

**robertocusato** · October 1st 2009, 11:10 PM

Hi,
can you help me with these two?

1. Find y' of y= √[(x-1)(x+2)^2

2. Use the quotient rule or the power rule to find y'.

if z = 2y^2 - 4y + 5, y = 6x - 5, and x = 2t, find dz/dt when t = 1.

Thanks folks!

**Amer** · October 2nd 2009, 10:14 AM

Originally Posted by robertocusato

Hi,
can you help me with these two?

1. Find y' of y= √[(x-1)(x+2)^2

2. Use the quotient rule or the power rule to find y'.

if z = 2y^2 - 4y + 5, y = 6x - 5, and x = 2t, find dz/dt when t = 1.

Thanks folks!

$y=((x-1)(x+2)^2)^{\frac{1}{2}}$

$y' = \left(\frac{1}{2}\right)((x-1)(x+2)^2)^{\frac{-1}{2}}\left(\frac{d((x-1)(x+2)^2)}{dx}\right)$

$y'=\left(\frac{1}{2}\right)((x-1)(x+2)^2)^{\frac{-1}{2}}\left((x+2)^2 + 2(x-1)(x+2)\right)$

$y'=\frac{(x+2)^2 + 2(x-1)(x+2)}{(2)\sqrt{(x-1)(x+2)^2}}$

show some work try other questions and present them if there is a mistake we will fix it

**HallsofIvy** · October 2nd 2009, 02:36 PM

Originally Posted by Amer

$y=\sqrt{(x-1)(x+2)^2}$

$y=(x+2)\sqrt{x-1}=(x+2)(x-1)^{\frac{1}{2}}$

Not quite. $y= |x+2|\sqrt{x-1}$ .

$y'= (x-1)^{\frac{1}{2}}\cdot \frac{d}{dx}(x+2) + (x+2)\cdot \frac{d}{dx}(x-1)^{\frac{1}{2}}$

$y'=(x-1)^{\frac{1}{2}}(1) + (x+2)\left(\frac{1}{2}\right)(1)(x-1)^{\frac{-1}{2}}$

show some work try other questions and present them if there is a mistake we will fix it

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