Find slope of the tangent line of $4\sqrt x + 2e^\frac {3x-12}{x+2}$ at $ x_0$ Find the slope and the equation of the tangent line to the
graph $y = f(x)$ at $x_0=4$,  $$4\sqrt x + 2e^\frac {3x-12}{x+2} $$
$$\lim_{h\to 0}\tfrac{4\sqrt {4+h} + 2e^\frac {12+3h-12}{4+h+2} - 10}{h} = \lim_{h\to 0}\frac{4\sqrt {4+h} + 2e^\frac {3h}{6+h} - 10}{h} $$ I am stuck at this part

From where you left by students's L'hospital
$$f'(4)=\lim_{h\rightarrow 0}\frac{\frac{2}{\sqrt{4+h}}+2e^{\frac{3h}{6+h}}\frac{3(6+h)-3h}{(6+h)^2}}{1}=\frac{2}{\sqrt{4}}+2.e^0.\frac{3.6}{6^2}=1+1=2$$
and the tangent line at $(4,10)$ is
$$\frac{y-10}{x-4}=2\implies y=2x+2.$$
