A hollow tube of length $L$, inner radius $R_{1}$ and outer radius $R_{2}$ is twisted by an angle $\alpha$ by applying a moment $M$. The angle $\alpha$ is a function of the inner radius $R_{1}$.
$$\alpha = \frac{2(1+v)L}{EJ}M \qquad \text{where} \qquad J = \frac{\pi}{2}(R_{2}^{4}-R_{1}^{4}) \tag{1}\label{eq1}$$
For a small change of $\Delta R_{1} \ll$ determine the resulting change $\Delta \alpha$ for the angle with the help of a linear approximation
Using the chain rule we can write the following equation:
$$\frac{\partial \alpha}{\partial R_{1}} = \frac{2 (1 + v) L M}{E}\frac{\frac{\pi}{2} 4 R_{1}^{3}}{J^{2}}\tag{2}\label{eq2}$$
given the fact that the chain rule says:
$$(f \circ u)' = \frac{d}{dx}f(u(x)) = f'(u(x)) \cdot u'(x) $$
What is exactly $f$, $f'$, $u$ and $u'$ in the equation (2) to make the transition from (1) to (2) more understandable?
$u$ and $f$ are as follows: \begin{align} u(R_1) &= \frac{\pi}{2}(R_{2}^{4}-R_{1}^{4})\\ &= J\\ f(J) &= \frac{1}{J}\\ f'(J) &= -\frac{1}{J^2}\\ u'(R_1) &= -\frac{\pi}{2}\times4R_1^3 \end{align} $\frac{2 (1 + v) L M}{E}$ is just a constant.