Derivative of Composition of Functions

$g'(\frac\pi6)=\frac{dg}{dx}(\frac\pi6)=[\frac{d}{dx}g(x)](\frac\pi6)=[\frac{d}{dx}f(sin(x))](\frac\pi6)=[f'(sin(x))\times\frac{d}{dx}sin(x)](\frac\pi6)=[f'(sin(x))\times cos(x)](\frac\pi6)$ $=f'(sin(\frac\pi6))\times cos(\frac\pi6)=f'(\frac12)\times\frac{\sqrt3}2=2\times\frac{\sqrt3}2=\sqrt3$

Note that $f'(\frac12)=2$ since the value of the derivative is the slope of the tangent, and the line equation given is in slope-intercept form, $y(x)=mx+y(0)$ ... Alternatively, the tangent passes through the $(0,1)$ and the $(\frac12,2)$ points and has a slope of $\frac{2-1}{\frac12-0}=2$ .

We have that $g(x)=f(\sin{x})$

Differentiate both sides with respect to $x$ (use the chain rule on the right-hand side) to get $\frac{dg}{dx}=\cos{x} \frac{d}{d(\sin{x})}f(\sin{x})$

Now, simply evaluate the above at $x=\frac{\pi}{6}$ to get $=\sqrt{3}$

using the fact that the instantaneous slope of $f(x)$ with respect to $x$ at $x=2$ is $2$ .