Hint : Recall about the formulas for differentiation or chain rule according to which differentiation of f(g(x)) = f'(g(x)). g'(x).

Solution :-
Given the function,
[tex] \rm y = \sqrt{x^2+4x-7}[/tex]
So, by differentiating both sides, we get:
[tex] \sf \implies \dfrac{dy}{dx} = \dfrac{d}{dx} ( \sqrt{ {x}^{2} + 4x - 7 } )[/tex]
We know that derivative of √x = 1/2√x.
[tex] \sf \implies \dfrac{dy}{dx} = \dfrac{1}{2\sqrt{ {x}^{2} + 4x - 7 }} .\dfrac{d}{dx} ( {x}^{2} + 4x - 7)[/tex]
We know that,
[tex] \boxed{\tt\dfrac{d}{dx} (Constant) = 0 }[/tex]
[tex] \boxed{\tt \dfrac{d}{dx} (a^n) = na^{n-1}}[/tex]
[tex] \sf \implies \dfrac{dy}{dx} = \dfrac{1}{2\sqrt{ {x}^{2} + 4x - 7 }} .(2x + 4)[/tex]
[tex] \sf \implies \dfrac{dy}{dx} = \dfrac{2(x + 2)}{2\sqrt{ {x}^{2} + 4x - 7 }}[/tex]
[tex]\implies\underline{\underline{ \sf \dfrac{dy}{dx} = \dfrac{x + 2}{\sqrt{ {x}^{2} + 4x - 7 }}}}[/tex]

This is the required derivative of given function.