Question: We are given the graphs of f and g below. Find (g o f)'(1) , i.e. the derivative of (g o f) at 1.

![enter image description here](https://i.stack.imgur.com/uHfjE.png)

my work is 

let $$g(x) = x^2 + 2$$

$$f(x) = \sqrt{x+2}$$

i'm stuck , is my work correct and f and g is correct to get (g o f) and (g o f)'

please look well in the graph and compare between my suggestion f,g , is they correct or no ? 
any help

Answer: 
Your functions (as written now) are incorrect. Observe that $g(x)$ is the translation of $x^2$ two units to the right, i.e. $g(x) = (x-2)^2 = x^2 - 4x + 4$. Though it is slightly unclear, $f(x)$ is most likely $e^x$ (a quick check on Mathematica shows this).
Now we can apply the chain rule. Since $g$ is differentiable at $f(1) = e$ and $f$ is differentiable at 1, we have
$$ (g \circ f)'(1) = g'(f(1))f'(1) = g'(e)f'(1) = (2e - 4)(e) = 2e^2 - 4e.$$
The composition $f \circ g$, of functions $f$ and $g$, is defined by the operation $ (f\circ g)(x) = f(g(x))$ for all $x$ in your domain. Since $g(x)= x^2+2$, and $f(x)=\sqrt{x+2}$, then $(f \circ g)(x) = f(g(x))=f(x^2+2)=\sqrt{(x^2+2)+2}=\sqrt{x^2+4}$

Then take the derivative to find $(f \circ g)'$ and substitute x=1 as per usual :)