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Wize University Calculus 2 Textbook > Multivariable Functions

Partial differentiation

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Partial Derviatives

There are 2 partial derivatives of the function f(x,y)f(x,y):
  • With respect to xx (treating yy like a constant): fx(x, y)=∂f∂x=lim⁡h→0f(x+h, y)−f(x, y)h\displaystyle f_x(x,\ y)=\frac{∂f}{∂x}=\lim_{h\rightarrow0}\frac{f(x+h,\ y)-f(x,\ y)}{h}
  • With respect to yy (trating xx like a constant): fy(x, y)=∂f∂y=lim⁡h→0f(x, y+h)−f(x, y)h\displaystyle f_y(x,\ y)=\frac{∂f}{∂y}=\lim_{h\rightarrow0}\frac{f(x,\ y+h)-f(x,\ y)}{h}
*If the partial derivatives fxf_x and fyf_y exist near (a,b)\left(a,b\right) and are continuous at (a,b)\left(a,b\right), then f(x,y)f\left(x,y\right) is differentiable at (a,b)\left(a,b\right)

Example
Find fx(0, 1)f_x(0,\ 1) and fy(0,1)f_y(0,1) if f(x,y)=y+y2+xy+x2 cos⁡x+exyf(x,y)=y+y^2+\frac{x}{y}+x^2\ \cos x+e^{xy}.
fxf_x (treat any y like a constant):
fx=0+0+1y+(2x)(cos⁡x)+(x2)(−sin⁡x)+exy(y)f_x=0+0+\frac{1}{y}+\left(2x\right)\left(\cos x\right)+\left(x^2\right)\left(-\sin x\right)+e^{xy}\left(y\right)
fx(0,1)=11+(0)(cos⁡0)+(0)(−sin⁡0)+e(0)(1)f_x\left(0,1\right)=\frac{1}{1}+\left(0\right)\left(\cos0\right)+\left(0\right)\left(-\sin0\right)+e^{\left(0\right)}\left(1\right)
fx(0,1)=2f_x\left(0,1\right)=2

fyf_y (treat any x like a constant):
fy=1+2y−xy2+0+exy(x)f_y=1+2y-\frac{x}{y^2}+0+e^{xy}\left(x\right)
fy(0,1)=1+2−(01)+e(0)(0)f_y\left(0,1\right)=1+2-\left(\frac{0}{1}\right)+e^{\left(0\right)}\left(0\right)
fy(0,1)=3f_y\left(0,1\right)=3

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Higher Order Partial Derviatives

The order of the partial derivatives is the number of partial derivatives taken of the function.

Second order partial derivatives of f(x,y)\bold{f\left(x,y\right)}
fxx=(fx)x=∂∂x(∂f∂x)=∂2f∂x2fyy=(fy)y=∂∂y(∂f∂y)=∂2f∂y2fxy=(fx)y=∂∂y(∂f∂x)=∂2f∂y∂xfyx=(fy)x=∂∂x(∂f∂y)=∂2f∂x∂y\begin{array}{|c|} \hline \displaystyle f_{xx}=(f_x)_x=\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial^2f}{\partial x^2}\\\\\hline \displaystyle f_{yy}=(f_y)_y=\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right)=\frac{\partial^2f}{\partial y^2}\\\\\hline \displaystyle f_{xy}=(f_x)_y=\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial^2f}{\partial y \partial x}\\\\\hline \displaystyle f_{yx}=(f_y)_x=\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)=\frac{\partial^2f}{\partial x \partial y}\\\\\hline \end{array}

*By Clairaut's Theorem, the mixed partial derivatives are the same (fxy=fyxf_{xy}=f_{yx})

*If fxx+fyy=0f_{xx}+f_{yy}=0 and the second partial derivatives are continuous on some domain, then f(x,y)f(x,y) is harmonic on that region

Example
Find fxx, fxy, fyyf_{xx},\ f_{xy},\ f_{yy} if f(x,y)=2x2y2+4x−3y2f\left(x,y\right)=2x^2y^2+4x-3y^2
fxx=∂ ∂x(4y2x+4)=4y2f_{xx}=\frac{\partial}{\ \partial x}\left(4y^2x+4\right)=4y^2
fyy=∂∂y(4x2y−6y)=4x2−6f_{yy}=\frac{\partial}{\partial y}\left(4x^2y-6y\right)=4x^2-6
fxy=∂∂y(4y2x+4)=8xyf_{xy}=\frac{\partial}{\partial y}\left(4y^2x+4\right)=8xy
*Note that fyx=fxy=8xyf_{yx}=f_{xy}=8xy

Practice: First Partial derivatives

Find fθ(0,π)f_{\theta}(0,\pi) and fϕ(0,π)f_{\phi} (0,\pi) if f(θ,ϕ)=eθϕcos⁡(ϕ) f\left({\theta},\phi\right)=e^{{\theta} \phi}\cos(\phi)\ .





Practice: Second Partial Derivatives

Find all the second-order partial derivatives of f(x,y)=xy+sin⁡(x2+y3).f(x,y)=xy+\sin(x^2+y^3).