Chapter 15 - Differentiation in Several Variables - 15.6 The Chain Rule - Exercises - Page 809: 32

$\frac{{\partial U}}{{\partial T}}{|_{\left( {T,U,V,W} \right) = \left( {1,1,2,4} \right)}} = - \frac{{2\ln 2}}{{1 + 2\ln 2}}$ $\frac{{\partial T}}{{\partial U}}{|_{\left( {T,U,V,W} \right) = \left( {1,1,2,4} \right)}} = - \frac{{1 + 2\ln 2}}{{2\ln 2}}$

We are given ${\left( {TU - V} \right)^2}\ln \left( {W - UV} \right) = 1$. Write $F\left( {T,U,V,W} \right) \equiv {\left( {TU - V} \right)^2}\ln \left( {W - UV} \right) - 1$. Then the partial derivatives are $\frac{{\partial F}}{{\partial T}} = 2U\left( {TU - V} \right)\ln \left( {W - UV} \right)$ $\frac{{\partial F}}{{\partial U}} = 2T\left( {TU - V} \right)\ln \left( {W - UV} \right) - \frac{{V{{\left( {TU - V} \right)}^2}}}{{W - UV}}$ $\frac{{\partial F}}{{\partial V}} = - 2\left( {TU - V} \right)\ln \left( {W - UV} \right) - \frac{{U{{\left( {TU - V} \right)}^2}}}{{W - UV}}$ $\frac{{\partial F}}{{\partial W}} = \frac{{{{\left( {TU - V} \right)}^2}}}{{W - UV}}$ At $\left( {T,U,V,W} \right) = \left( {1,1,2,4} \right)$ we obtain $\frac{{\partial F}}{{\partial T}}{|_{\left( {T,U,V,W} \right) = \left( {1,1,2,4} \right)}} = - 2\ln 2$ $\frac{{\partial F}}{{\partial U}}{|_{\left( {T,U,V,W} \right) = \left( {1,1,2,4} \right)}} = - 1 - 2\ln 2$ $\frac{{\partial F}}{{\partial V}}{|_{\left( {T,U,V,W} \right) = \left( {1,1,2,4} \right)}} = - \frac{1}{2} + 2\ln 2$ $\frac{{\partial F}}{{\partial W}}{|_{\left( {T,U,V,W} \right) = \left( {1,1,2,4} \right)}} = \frac{1}{2}$ 1. Evaluate $\frac{{\partial U}}{{\partial T}}$ Suppose $U$ is defined implicitly as a function of $T$, $V$ and $W$ by the equation $F\left( {T,U,V,W} \right) = 0$. Using the Chain Rule we differentiate $F$ with respect to $T$: $\frac{{\partial F}}{{\partial T}}\frac{{\partial T}}{{\partial T}} + \frac{{\partial F}}{{\partial U}}\frac{{\partial U}}{{\partial T}} + \frac{{\partial F}}{{\partial V}}\frac{{\partial V}}{{\partial T}} + \frac{{\partial F}}{{\partial W}}\frac{{\partial W}}{{\partial T}} = 0$ Since $\frac{{\partial T}}{{\partial T}} = 1$, ${\ \ }$ $\frac{{\partial V}}{{\partial T}} = 0$, ${\ \ }$ $\frac{{\partial W}}{{\partial T}} = 0$, so $\frac{{\partial F}}{{\partial T}} + \frac{{\partial F}}{{\partial U}}\frac{{\partial U}}{{\partial T}} = 0$ $\frac{{\partial U}}{{\partial T}} = - \frac{{\partial F}}{{\partial T}}/\frac{{\partial F}}{{\partial U}}$ At $\left( {T,U,V,W} \right) = \left( {1,1,2,4} \right)$ we obtain $\frac{{\partial U}}{{\partial T}}{|_{\left( {T,U,V,W} \right) = \left( {1,1,2,4} \right)}} = - \frac{{\partial F}}{{\partial T}}{|_{\left( {T,U,V,W} \right) = \left( {1,1,2,4} \right)}}/\frac{{\partial F}}{{\partial U}}{|_{\left( {T,U,V,W} \right) = \left( {1,1,2,4} \right)}}$ $\frac{{\partial U}}{{\partial T}}{|_{\left( {T,U,V,W} \right) = \left( {1,1,2,4} \right)}} = - \frac{{2\ln 2}}{{1 + 2\ln 2}}$ 2. Evaluate $\frac{{\partial T}}{{\partial U}}$ Suppose $T$ is defined implicitly as a function of $U$, $V$ and $W$ by the equation $F\left( {T,U,V,W} \right) = 0$. Using the Chain Rule we differentiate $F$ with respect to $U$: $\frac{{\partial F}}{{\partial T}}\frac{{\partial T}}{{\partial U}} + \frac{{\partial F}}{{\partial U}}\frac{{\partial U}}{{\partial U}} + \frac{{\partial F}}{{\partial V}}\frac{{\partial V}}{{\partial U}} + \frac{{\partial F}}{{\partial W}}\frac{{\partial W}}{{\partial U}} = 0$ Since $\frac{{\partial U}}{{\partial U}} = 1$, ${\ \ }$ $\frac{{\partial V}}{{\partial U}} = 0$, $\frac{{\partial W}}{{\partial U}} = 0$, so $\frac{{\partial F}}{{\partial T}}\frac{{\partial T}}{{\partial U}} + \frac{{\partial F}}{{\partial U}} = 0$ $\frac{{\partial T}}{{\partial U}} = - \frac{{\partial F}}{{\partial U}}/\frac{{\partial F}}{{\partial T}}$ At $\left( {T,U,V,W} \right) = \left( {1,1,2,4} \right)$ we obtain $\frac{{\partial T}}{{\partial U}}{|_{\left( {T,U,V,W} \right) = \left( {1,1,2,4} \right)}} = - \frac{{\partial F}}{{\partial U}}{|_{\left( {T,U,V,W} \right) = \left( {1,1,2,4} \right)}}/\frac{{\partial F}}{{\partial T}}{|_{\left( {T,U,V,W} \right) = \left( {1,1,2,4} \right)}}$ $\frac{{\partial T}}{{\partial U}}{|_{\left( {T,U,V,W} \right) = \left( {1,1,2,4} \right)}} = - \frac{{1 + 2\ln 2}}{{2\ln 2}}$