Find $\displaystyle \frac{dy}{dx}$ of the curve $\displaystyle y=\sqrt{4x-7}-\frac{2}{3}(x-4)$ and show that $\displaystyle \frac{d^2y}{dx^2} = \frac{-4}{(4x-7)\sqrt{4x-7}}$
First, this is the not the result you posted earlier.
Secondly, it should be $\displaystyle \frac{2}{(4x-7)^{0.5}}-\frac{2}{3}$ not $\displaystyle \frac{2}{(4x-7)^{{\color{red}-}0.5}}-\frac{2}{3}$.
And thirdly, I'd leave it as $\displaystyle =\frac{2}{(4x-7)^{0.5}}-\frac{2}{3}$ because this is easier to differentiate (to get the second derivative) than the final form you have given.