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
 A: You can do this by using the direct calculus of derivatives, without passing through the incremental ratio.
The equation of the tangent line $r$ at $x = x_0$ for a certain $y = f(x)$ is given by
$$r: f(x_0) + f'(x_0)\cdot (x-x_0)$$
And its slope is just $f'(x_0)$.
The slope is given by $f'(x_0)$ where $x_0 = 4$ in your cases. Thence
$$f(x) = 4\sqrt{x} + 2e^{\frac{3x-12}{x+2}} \longrightarrow f'(x) = \frac{2}{\sqrt{x}} + 2 e^{\frac{3 x-12}{x+2}} \left(\frac{3}{x+2}-\frac{3 x-12}{(x+2)^2}\right)$$
Or arranging it a bit
$$f'(x) = \frac{2}{\sqrt{x}}+\frac{36 e^{\frac{3 (x-4)}{x+2}}}{(x+2)^2}$$
Now it's easy:
$$f'(4) = 2$$
A: $$\lim_{h\to 0}\frac{4\sqrt{4+h}-8}{h}+\lim_{h\to 0}\frac{2e^{\tfrac{3h}{6+h}}-2}{h}$$$$=4\lim_{h\to 0}\frac{\sqrt{4+h}-\sqrt4}{h}+2\lim_{h\to0}\frac{e^{\tfrac{3h}{6+h}}-1}{h}$$Rationalising the first limit and adjusting the denominator of the second limit, $$ =4\lim_{h\to 0}\frac{h}{h\cdot(\sqrt{4+h}+\sqrt4)}+2\times 3\lim_{h\to0}\frac{e^{\tfrac{3h}{6+h}}-1}{\frac{3h}{6+h}\cdot (6+h)}$$ Using direct substitution of $h=0$ on the first limit and the formula $\lim \limits_{U\to 0}\dfrac{e^U-1}{U}=1$ on the second limit,$$=4\cdot\frac{1}{2+2}+\frac{2\times3}{6\times 1}=2.$$
A: 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
$$\frac{y-10}{x-4}=2\implies y=2x+2.$$
