## Calculus: Early Transcendentals 8th Edition

$\dfrac{\partial R}{\partial R_{1}}=[\dfrac{R}{R_{1}}]^2$
We need to apply $\dfrac{\partial}{\partial R_{1}}$ to both sides of the given equation. Apply chain rule in the LHS equation as R is a function of $R_{1}.$ and in the RHS $R_{2},$ and $R_{3}$ have to be treated as constants. $\dfrac{\partial}{\partial R_{1}}[R^{-1}]=\dfrac{\partial}{\partial R_{1}}[R_{1}^{-1}]+0+0$ This implies that $-R^{-2}\times \dfrac{\partial R}{\partial R_{1}}=-R_{1}^{-2}$ Hence, we have $\dfrac{\partial R}{\partial R_{1}}=[\dfrac{R}{R_{1}}]^2$