5.6 Higher Order Derivatives

When we differentiate a function \(f(x,y)\) twice, we produce its second order derivatives. There derivatives are denoted by \(\hspace {0.2cm} \dfrac {\partial ^2f}{\partial x^2},\hspace {0.3cm} \dfrac {\partial ^2f}{\partial y^2}, \hspace {0.4cm} \dfrac {\partial ^2f}{\partial x\partial y}\hspace {0.2cm}\) or \(\hspace {0.2cm} f_{xx},\hspace {0.2cm} f_{yy},\hspace {0.2cm} f_{xy}\).

These derivatives are defined as

\begin {align*} f_{xx} & = \Big (f_x\Big )_x = \dfrac {\partial }{\partial x}\Bigg [\dfrac {\partial f}{\partial x}\Bigg ]\\\\ f_{yy} & = \Big (f_y\Big )_y = \dfrac {\partial }{\partial y}\Bigg [\dfrac {\partial f}{\partial y}\Bigg ]\\\\ f_{yx} & = \Big (f_y\Big )_x = \dfrac {\partial }{\partial x}\Bigg [\dfrac {\partial f}{\partial y}\Bigg ] = \dfrac {\partial ^2f}{\partial x \partial y}\\\\ f_{xy} & = \Big (f_y\Big )_y = \dfrac {\partial }{\partial y}\Bigg [\dfrac {\partial f}{\partial x}\Bigg ] = \dfrac {\partial ^2f}{\partial y \partial x}\\\\ \end {align*}

Example 5.6.1.

1.: Find all the second partial derivatives of \[f(x,y) = x\cos y + ye^x\]

Solution. \(\dfrac {\partial f}{\partial x} = \cos y + ye^x\hspace {0.2cm}, \hspace {0.3cm} \dfrac {\partial f}{\partial y} = -x\sin y + e^x\)

\(f_{xx} = \dfrac {\partial }{\partial x}\big [\cos y + ye^x\big ] = ye^x\)

\(f_{xy} = \dfrac {\partial }{\partial y}(\cos y + ye^x) = -\sin y + e^x\)

\(f_{yy} = \dfrac {\partial }{\partial y}\big [-x\sin y + e^x\big ] = -x\cos y\)

\(f_{yx} = \dfrac {\partial }{\partial x}\big [-x\sin y + e^x\big ] = -\sin y + e^x\)

5.6.1 The mixed Derivative Theorem

If \(f(x,y)\) and its partial derivatives \(f_x, \hspace {0.2cm} f_y, \hspace {0.2cm} f_{xy}, \hspace {0.2cm} f_{yx}\) are defined in a region containing a point \((a,b)\) and are all continuous at \((a,b)\) then \(f_{xy}(a,b) = f_{yx}(a,b)\).

Example 5.6.2.

1.: Find \(z_{xy}\) if \(z = f(u,v), \hspace {0.3cm} u = x + y,\hspace {0.3cm} v =xy\) and \(f\) its partial derivatives are continuous.
2.: Find \(\dfrac {d^2w}{dt^2}\) if \(w = f(x,y),\hspace {0.3cm} x = e^t,\hspace {0.3cm} y = 2t - 1\) and \(w\) and its partial derivatives are all continuous.

Solution.

Part 1

\begin {align*} z_x = \dfrac {\partial f}{\partial x} & = \frac {\partial f}{\partial u}\cdot \frac {\partial u}{\partial x} + \frac {\partial f}{\partial v}\cdot \frac {\partial v}{\partial x}\\\\ & = \frac {\partial f}{\partial u} + y\dfrac {\partial f}{\partial v}\\\\ & = f_u + yf_v \end {align*}

\begin {align*} z_{xy} & = \frac {\partial ^2f}{\partial y \partial x} = \frac {\partial }{\partial y}\big [f_u + y f_v\big ]\\\\ & = \frac {\partial }{\partial y}\big [f_u\big ] + f_v + y \dfrac {\partial }{\partial y}\big [f_v\big ]\\\\ & = f_{uu}\frac {\partial u}{\partial y} + f_{uv}\frac {\partial v}{\partial y} +f_v + y\Bigg [f_{vu}\frac {\partial u}{\partial y} + f_{vv}\frac {\partial v}{\partial y}\Bigg ]\\\\ & = f_{uu} + xf_{uv} + f_v + yf_{vu} + xyf_{vv} \end {align*}

But \(f_{uv} = f_{vu}\) since \(f\) is its partial derivatives are continuous

\[z_{xy} = f_{uu} + (x+y)f_{uv} + xyf_{vv} + f_v\]

But \(x + y = u, \hspace {0.3cm} xy = v \) so \[z_{xy} = f_{uu} + uf_{uv} + vf_{vv} + f_v\]

Part 2

\(w = f(x,y),\hspace {0.4cm} y\) and \(x\) functions of \(t\)

\[\dfrac {d w}{dt} = \dfrac {\partial f}{\partial x}\cdot \dfrac {dx}{dt} + \dfrac {\partial f}{\partial y}\cdot \dfrac {dy}{dt} = e^t\dfrac {\partial f}{\partial x} + 2\dfrac {\partial f}{\partial y}\]

\begin {align*} \dfrac {d^2w}{dt^2} & = \dfrac {d}{dt}\Bigg [ e^t\dfrac {\partial f}{\partial x} + 2\dfrac {\partial f}{\partial y}\Bigg ]\\ & = \dfrac {d}{dt}\Bigg [ e^t\dfrac {\partial f}{\partial x}\Bigg ] + 2\dfrac {d}{dt}\Bigg [\dfrac {\partial f}{\partial y}\Bigg ]\\\\ & = e^t\dfrac {\partial f}{\partial x} + e^t\dfrac {\partial }{\partial t}\Bigg [\dfrac {\partial f}{\partial x}\Bigg ] + 2 \Bigg [\dfrac {\partial ^f}{\partial x\partial y}\cdot \dfrac {dx}{dt} + \dfrac {\partial ^2f}{\partial y^2}\cdot \dfrac {dy}{dt}\Bigg ]\\\\ & = e^t\dfrac {\partial f}{\partial x} + e^t \Bigg [\dfrac {\partial ^f}{\partial x^2}\cdot \dfrac {dx}{dt} + \dfrac {\partial ^f}{\partial y \partial x}\cdot \dfrac {dy}{dt}\Bigg ] + 2 \Bigg [\dfrac {\partial ^2f}{\partial x\partial y}\cdot \dfrac {dx}{dt} + \dfrac {\partial ^2f}{\partial y^2}\cdot \dfrac {dy}{dt}\Bigg ]\\\\ & = e^t\dfrac {\partial f}{\partial x} + e^t\Bigg [e^t\dfrac {\partial ^2f}{\partial x^2} + 2\dfrac {\partial ^2f}{\partial y \partial x}\Bigg ] + 2\Bigg [e^t\dfrac {\partial ^2f}{\partial x\partial y} + 2 \dfrac {\partial ^2f}{\partial y^2}\Bigg ]\\\\ & = e^t\frac {\partial f}{\partial x} + e^{2t}\dfrac {\partial ^2f}{\partial x^2} + 2e^t\dfrac {\partial ^2f}{\partial x\partial y} + 4\dfrac {\partial ^2f}{\partial y^2}\\\\ & = e^t\dfrac {\partial f}{\partial x} + e^{2t}\dfrac {\partial ^2f}{\partial x^2} + 4e^t\dfrac {\partial ^2f}{\partial x \partial y} + 4\dfrac {\partial ^2f}{\partial y^2}\\\\ & = e^tf_x + e^{2t}f_{xx} + 4e^{t}f_{yx} + 4f_{yy}\\\\ \end {align*}

An extension of the theorem above leads to \(\dfrac {\partial ^3f}{\partial x^2 \partial y} = \dfrac {\partial ^3f}{\partial y \partial x^2}\).

Example 5.6.3.

Calculate \(\dfrac {\partial ^5}{\partial x^2\partial y^3}\big [x\sin y + e^y\big ]\)

Solution.

By the theorem, the order of differentiation does not matter. Let \(f(x,y) = x\sin y + e^y\) \begin {align*} \dfrac {\partial ^5f}{\partial x^2\partial y^3} & = \dfrac {\partial ^5f}{\partial y^3\partial x^2}= \dfrac {\partial ^3}{\partial ^3}\Bigg [\dfrac {\partial ^2f}{\partial x^2}\Bigg ]\\\\ & = \dfrac {\partial ^3}{\partial y^3}\Bigg [\dfrac {\partial }{\partial x}\Bigg \{\dfrac {\partial f}{\partial x}\Bigg \}\Bigg ]\\\\ & = \dfrac {\partial ^3}{\partial y^3}\Bigg [\dfrac {\partial }{\partial x}\big (\sin y\big )\Bigg ]\\\\ & = 0\\\\ \end {align*}

5.6.2 The Implicit Function Theorem

1.: If \(F\) is defined on a disc containing \((a,b)\), where \(F(a,b) = 0,\hspace {0.2cm} F_y(a,b) \neq 0, \hspace {0.2cm} F_x\) and \(F_y\) are continuous on the disc, then the equation \(F(x,y) = 0\) defines \(y\) as a function of \(x\) near a point \((a,b)\) and the derivative of this function is \[\dfrac {dy}{dx} = \dfrac {\dfrac {-\partial F}{\partial x}}{\dfrac {\partial F}{\partial y}} = \dfrac {-F_x}{F_y}\]
2.: If \(F\) is defined within a sphere containing \((a,b,c)\) where \(F(a,b,c) = 0,\hspace {0.2cm} F_z(a,b,c) \neq 0,\hspace {0.2cm}\\ F_x,\hspace {0.2cm} F_y\) and \(F_z\) are continuous inside the sphere then the equation \(F(x,y) = 0\) defines \(z\) as a function of \(x\) and \(y\) near \((a,b,c)\) and this function is differentiable with partial derivatives \[\dfrac {\partial z}{\partial x} = \dfrac {-F_x}{F_z} \hspace {0.3cm}\text {and}\hspace {0.3cm} \dfrac {\partial z}{\partial y} = \frac {-F_y}{F_z}\]

Example 5.6.4.

1.: Find \(y'\) if \(x^3 + y^3 = 6xy\)
2.: Find \(\dfrac {\partial z}{\partial x}\) and \(\dfrac {\partial z}{\partial y}\) if \(x^3 + y^3 + z^3 + 6xyz = 1\).

Solution.

1.

Let \(F(x,y) = x^3 + y^3 - 6xy = 0\). Thus, \(F_y (x,y) = 3y^2 - 6x\). So by IFT we get \[y' = \dfrac {dy}{dx} = \dfrac {-F_x}{F_y} = \dfrac {-(3x^2 - 6y)}{3y^2 - 6x} = \dfrac {2y - x^2 }{y^2 - 2x}\]

2.

Let \(F(x,y,z) = x^3 + y^3 + z^3 + 6xyz - 1 = 0\) \[\dfrac {\partial F}{\partial z} = 3z^2 + 6xy \hspace {0.3cm} , \hspace {0.3cm} \dfrac {\partial F}{\partial x} = 3x^2 + 6yz\hspace {0.3cm} , \hspace {0.3cm} \dfrac {\partial F}{\partial y} = 3z^2 + 6xz\]

(a): \(\displaystyle {\dfrac {\partial z}{\partial x} = \dfrac {-\dfrac {\partial F}{\partial x}}{\dfrac {\partial F}{\partial z}} = \dfrac {-(3x^2 + 6yz)}{3z^2 + 6xy}}\)
(b): \(\displaystyle {\dfrac {\partial z}{\partial y} = \dfrac {-\dfrac {\partial F}{\partial y}}{\dfrac {\partial F}{\partial z}} = \dfrac {-(3y^2 + 6xz)}{3z^2 + 6xy}}\)

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