Let $y = \sqrt{4-3x}$.
The problem is to find $\dfrac{d}{dx} \sqrt{4-3x}$, i.e. to find $\dfrac{dy}{dx}$.
Let $u=4-3x$. Then $y=\sqrt{u}$.
Then we have $ \frac{dy}{du} = ? or \frac{1}{2}(4-3x)^\frac{-1}{2},\qquad \text{and}\qquad \frac{du}{dx} = -3. $
Therefore $ \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} = $
the right answer is $\dfrac{-3}{2\sqrt{4-3x}}$
Can you please help me out? thanks