Suppose \(z=f(x,y).\) The partial derivatives of \(f\) with respect to \(x\) is denoted by \[\frac{\partial z}{\partial x}\quad\mathrm{or~}\quad f_x(x,y)\mathrm{~or~}\frac{\partial f}{\partial x}\]
and is the function obtained by differentiating \(f\) with respect to \(x\), treating \(y\) as a constant. The partial derivative of \(f\) with respect to \(y\) is denoted by \[\frac{\partial z}{\partial y}\quad\mathrm{or}\quad f_y(x,y)\mathrm{~or~}\frac{\partial f}{\partial y}\]
and is the function obtained by differentiating \(f\) with respect to \(y\), treating \(x\) as a constant.
Computation of Partial Derivatives
No new rules are needed for the computation of partial derivatives.
Example.
Find the partial derivatives \(f_x\mathrm{~and~}f_y,\mathrm{if~}f(x,y)=x^2+2xy^2+\frac{2y}{3x}\)
Solution.
\(\begin{aligned}f_x(x,y)&=2x+2(1)y^2+\frac23y(-x^{-2})=2x+2y^2-\frac{2y}{3x^2}\\f_y(x,y)&=0+2x(2y)+\frac23(1)(x^{-1})=4xy+\frac2{3x}\end{aligned}\)
Example.
Find the partial derivatives \(f_x\mathrm{~and~}f_y,\mathrm{if~}f(x,y)=xe^{-2xy}.\)
Solution.
\(f_x(x,y)=x(-2ye^{-2xy})+e^{-2xy}=e^{-2xy}(1-2xy)\)
\(f_y(x,y)=x(-2xe^{-2xy})=-2x^2e^{-2xy}\)
Second-Order Partial Differentiation
If \(z = f (x, y)\), the partial derivative of \(f_x\) with respect to \(x\) is
The partial derivative of \(f_x\) with respect to \(y\) is
\(f_{xy}=\left(f_x\right)_y\quad\mathrm{or}\quad\frac{\partial^2z}{\partial y\partial x}=\frac\partial y{\left(\frac{\partial z}{\partial x}\right)}\mathrm{~or~}\quad\frac{\partial^2f}{\partial yx}=\frac\partial y{\left(\frac{\partial f}{\partial x}\right)}.\)
The partial derivative of \(f_y\) with respect to \(x\) is
\(f_{yx}=\left(f_y\right)_x\quad\mathrm{or}\quad\frac{\partial^2z}{\partial x\partial y}=\frac\partial x\left(\frac{\partial z}{\partial y}\right)\mathrm{~or~}\quad\frac{\partial^2f}{\partial xy}=\frac\partial x{\left(\frac{\partial f}{\partial y}\right)}.\)
The partial derivative of \(f_y\) with respect to \(y\) is
\(f_{yy}=\left(f_y\right)_y\mathrm{~or~}\quad\frac{\partial^2z}{\partial y^2}=\frac\partial y{\left(\frac{\partial z}{\partial y}\right)}\mathrm{~or~}\quad\frac{\partial^2f}{\partial y^2}=\frac\partial y{\left(\frac{\partial f}{\partial y}\right)}.\)
The two partial derivatives \(f_{xy}\) and \(f_{yx}.\) are sometimes called mixed second-order partial derivatives of \(f\). If both mixed second-order partial derivatives are continuous (on some disk), then they are equal, i.e., mixed partial derivatives are equal \(f_{xy}=f_{yx}.\)
Example.
Compute the four second-order partial derivatives of the function \(f(x,y)=xy^3+5xy^2+2x+1.\)
Solution.
\(f_x=y^3+5y^2+2.\) Then it follows that
\(f_{xx}=0\mathrm{~and~}f_{xy}=3y^2+10y\)
\(\begin{aligned}f_y&=3xy^2+10xy.\text{ Then, it follows that}\\f_{yy}&=6xy+10x\mathrm{~and~}f_{yx}=3y^2+10y\end{aligned}\)
Example.
Compute the four second-order partial derivatives of the function \(f(x,y)=x^2ye^x.\)
Solution.
\(f_x=y(2xe^x+x^2e^x)\), Then it follows that
\(\begin{aligned} f_{xx} & =y[2(e^x+xe^x)+2xe^x+x^2e^x] \\ & =y(4xe^x+2e^x+x^2e^x) \end{aligned}\)
and \(f_{xy}=2xe^x+x^2e^x\).
\(f_y=x^2e^x\). Then, it follows that
\(f_{yy}=0\) and \(f_{yx}=2xe^x+x^2e^x\).