Giải thích các bước giải:
a.Ta có:
$y=\dfrac{3}{\sqrt{3x-2}}$
$\to y'=(\dfrac{3}{\sqrt{3x-2}})'$
$\to y'=3\left(\left(3x-2\right)^{-\dfrac{1}{2}}\right)'\:$
$\to y'=3\cdot (3x-2)'\cdot \dfrac{-1}{2}\left(\left(3x-2\right)^{-\dfrac{1}{2}-1}\right)$
$\to y'=-\dfrac{9}{2(3x-2)^{\dfrac32}}$
b.Ta có:
$y=\dfrac{2x+3}{\sqrt{x+1}}$
$\to y=\dfrac{2(x+1)+1}{\sqrt{x+1}}$
$\to y=2\sqrt{x+1}+\dfrac{1}{\sqrt{x+1}}$
$\to y=2(x+1)^{\dfrac12}+(x+1)^{-\dfrac12}$
$\to y'=2\cdot \dfrac12 \cdot (x+1)^{\dfrac12-1}-\dfrac12\cdot (x+1)^{-\dfrac12-1}$
$\to y'=\dfrac{2x+1}{2\left(x+1\right)^{\dfrac{3}{2}}}$
c.Ta có:
$y=\dfrac{\sqrt{2x-2}}{3x+1}$
$\to y'=(\dfrac{\sqrt{2x-2}}{3x+1})'$
$\to y'=\dfrac{\left(\sqrt{2x-2}\right)'\left(3x+1\right)-\left(3x+1\right)'\sqrt{2x-2}}{\left(3x+1\right)^2}$
$\to y'=\dfrac{\dfrac{1}{\sqrt{2x-2}}\left(3x+1\right)-3\sqrt{2x-2}}{\left(3x+1\right)^2}$
$\to y'=\dfrac{-3x+7}{\left(3x+1\right)^2\sqrt{2x-2}}$
