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
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