Differentiating a Function of Three Variables Where Each Variable Is a Function of Four Variables

Find \(\dfrac{\partial f}{\partial u},\) \(\dfrac{\partial f}{\partial v},\) \(\dfrac{\partial f}{\partial w},\) \(\dfrac{\partial f}{\partial t}\) for the function \(f(x,y,z)=x^{2}+y^{2}+z^{2},\) where \(x=uvwt\), \(y=e^{u+v+w+t}\), and \(z=u+2v+3w+4t\).

Solution We use an extension of Chain Rule II. \[ \begin{array}{lllllll} \dfrac{\partial f}{\partial u} &=&\dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial u}+\dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial u}+ \dfrac{\partial f}{\partial z}\dfrac{\partial z}{\partial u} & \dfrac{ \partial f}{\partial v}&=&\dfrac{\partial f}{\partial x}\dfrac{\partial x}{ \partial v}+\dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial v}+\dfrac{ \partial f}{\partial z}\dfrac{\partial z}{\partial v} \nonumber\\ &=&(2x)(vwt)+(2y)(e^{u+v+w+t})+(2z)(1) & &=&(2x)(uwt)+(2y)(e^{u+v+w+t})+(2z)(2) \nonumber \\ &=&2uv^{2}w^{2}t^{2}+2e^{2(u+v+w+t)}+2(u+2v+3w+4t) & &=&2u^{2}vw^{2}t^{2}+2e^{2(u+v+w+t)}+4(u+2v+3w+4t) \nonumber \\[1pc] \dfrac{\partial f}{\partial w} &=&\dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial w}+\dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial w}+ \dfrac{\partial f}{\partial z}\dfrac{\partial z}{\partial w} & \dfrac{\partial f}{ \partial t}&=&\dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial t}+\dfrac{ \partial f}{\partial y}\dfrac{\partial y}{\partial t}+\dfrac{\partial f}{ \partial z}\dfrac{\partial z}{\partial t} \nonumber \\ &=&(2x)(uvt)+(2y)(e^{u+v+w+t})+(2z)(3) & &=&(2x)(uvw)+(2y)(e^{u+v+w+t})+(2z)(4) \nonumber \\ &=&2u^{2}v^{2}wt^{2}+2e^{2(u+v+w+t)}+6(u+2v+3w+4t) & &=&2u^{2}v^{2}w^{2}t+2e^{2(u+v+w+t)}+8(u+2v+3w+4t) \end{array} \]

Again, notice that the partial derivatives of \(f\) in Example 5 are expressed in terms of \(u, v, w\), and \(t\) alone.