Answer
$\frac{d}{dx} f(2sin(x))|_{x=\frac{\pi}{6}}=-\frac{5\sqrt{3}}{2}$
Work Step by Step
We begin by using the graph to find $f(1)$ and $f'(1)$:
We can approximate $f(x)$ from $(0,2)$ by noticing that $f(x)$ is a line at this interval and crosses the points $(0,5)$ and $(2,0)$.
The slope of this line is therefore: $\frac{\Delta y}{\Delta x} = \frac{0-5}{2-0} = \frac{-5}{2}$
Using point-slope form with the point $(0,5)$ and slope $\frac{4}{3}$, the equation of the line is $y-5 = -\frac{5}{2}(x-0)$
Therefore, $f(1) = \frac{5}{2}$ and $f'(-1) = \frac{-5}{2}$ (as $f'(-1)$ is just the slope of the line.
We apply the power rule and then the chain rule:
$\frac{d}{dx} f(2sin(x))|_{x=\frac{\pi}{6}} = [f'(2sin(x)) * \frac{d}{dx}[2sin(x)]]|_{x=\frac{\pi}{6}} = [f'(2sin(x)) * (2cos(x))]|_{x=\frac{\pi}{6}} =[f'(2sin(\frac{\pi}{6})) * (2cos(\frac{\pi}{6}))]|_{x=\frac{\pi}{6}} = f'(2*\frac{1}{2}) * (2*\frac{\sqrt{3}}{2}) = f'(1) * (\sqrt{3}) = -\frac{5}{2} * \sqrt{3} = -\frac{5\sqrt{3}}{2}$
