## Calculus 8th Edition

$\dfrac{\partial R}{\partial R_{1}}=[\dfrac{R}{R_{1}}]^2$
We will have to use $\dfrac{\partial}{\partial R_{1}}$ to both sides of the given equation. Now, we will use chain rule in the LHS equation as R is a function of $R_{1}.$ Consider $R_{2},$ and $R_{3}$ keep treating as constants. $\dfrac{\partial}{\partial R_{1}}[R^{-1}]=\dfrac{\partial}{\partial R_{1}}[R_{1}^{-1}]+0+0$ This implies that $-[R]^{-2}\times \dfrac{\partial R}{\partial R_{1}}=-R_{1}^{-2}$ Thus, we get $\dfrac{\partial R}{\partial R_{1}}=[\dfrac{R}{R_{1}}]^2$