1. ## Derivatives

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!

2. 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!

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

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

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

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

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

3. Originally Posted by Amer
$\displaystyle y=\sqrt{(x-1)(x+2)^2}$

$\displaystyle y=(x+2)\sqrt{x-1}=(x+2)(x-1)^{\frac{1}{2}}$
Not quite. $\displaystyle y= |x+2|\sqrt{x-1}$.

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

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

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