5.5 The Chain Rule

The Chain Rule for functions of two variables

1.: If \(w = f(x,y)\) where \(x\) and \(y\) are differentiable functions of \(t\), then \[\dfrac {d w}{d t} = \frac {\partial f}{\partial x}\cdot \frac {dx}{dt} + \frac {\partial f}{\partial y}\cdot \frac {dy}{dt}\]
2.: If \(w = f(x,y)\) , where \(x\) and \(y\) are functions of \(r\) and \(s\), then \begin {align*} \dfrac {\partial f}{\partial r} & = \frac {\partial f}{\partial x}\cdot \frac {\partial x}{\partial r} + \frac {\partial f}{\partial y}\cdot \frac {\partial y}{\partial r}\\\\ \dfrac {\partial f}{\partial s} & = \frac {\partial f}{\partial x}\cdot \frac {\partial x}{\partial s} + \frac {\partial f}{\partial y}\cdot \frac {\partial y}{\partial s}\\\\ \end {align*}

Example 5.5.1.

1.: Use the chain rule to find the derivative of \(f(x,y) = xy\) with respect to \(t\) along the path \(x = \cos t, \hspace {0.2cm} y = \sin t\). What is the value of the derivative of \(t = \dfrac {\pi }{2}\)?

Solution. \begin {align*} \dfrac {d f}{d t} & = \frac {\partial f}{\partial x}\cdot \frac {dx}{dt} + \frac {\partial f}{\partial y}\cdot \frac {dy}{dt}\\ & = y(-\sin t) + x(\cos t) = -y \sin t + x\cos t\\ & = -\sin ^2 t + \cos ^2 t\\ & = \cos 2t\\ \frac {d f\Big (\dfrac {\pi }{2}\Big )}{dt} & = \cos 2\Big (\dfrac {\pi }{2}\Big ) = \cos \pi = 1\\\\ \end {align*}

5.5.1 The Chain Rule of Functions of Three Variations

If \(w = f(x,y,z)\) and its partial derivatives are continuous and \(x = x(t), \hspace {0.2cm} y = y(t), \hspace {0.2cm} z = z(t)\) are differentiable functions of \(t\), then \[\dfrac {d f}{d t} = \frac {\partial f}{\partial x}\cdot \frac {dx}{dt} + \frac {\partial f}{\partial y}\cdot \frac {dy}{dt} + \frac {\partial f}{\partial z}\cdot \frac {dz}{dt}\]

Example 5.5.2.

Find the value of \(\dfrac {df}{dt}\) at \(t = 0\) if \(f(x,y,z) = xy + z, \hspace {0.3cm} x = \cos t, \hspace {0.3cm} y = \sin t, \hspace {0.3cm} z = t\)

\begin {align*} \frac {d f}{d t} & = y (-\sin t) + x(\cos t) + 1(1)\\ & = \cos t + 1\\\\ \dfrac {d f}{d t}\Bigg |_{t = 0} & = 2\\\\ \end {align*}

Example 5.5.3.

Find \(\dfrac {\partial f}{\partial r}\) and \(\dfrac {\partial f}{\partial s}\) if \[f(x,y,z) = x + 2y + z^2, \hspace {0.4cm} x = \dfrac {r}{s},\hspace {0.4cm} y = r^2 + \ln s, \hspace {0.4cm} z = 2r\]

Solution. \begin {align*} \dfrac {\partial f}{d r} & = \frac {\partial f}{\partial x}\cdot \frac {\partial x}{dr} + \frac {\partial f}{\partial y}\cdot \frac {\partial y}{dr} + \frac {\partial f}{\partial z}\cdot \frac {\partial z}{dr}\\ & = 1\Big (\dfrac {1}{s}\Big ) + 2(2r) + 2z(2)\\ & = \dfrac {1}{s} + 4r + 4z\\ & = \dfrac {1}{s} + 4r + 4(2r)\\ \implies \hspace {0.4cm} \dfrac {\partial f}{\partial r} & = \dfrac {1}{s} + 12r\\\\ \dfrac {\partial f}{\partial s} & = \dfrac {2}{s} - \dfrac {r}{s^2}\\\\ \end {align*}

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