14.3 Partial Derivatives (Wk 6)

Define the partial derivatives $f_x(a,b)$ and $f_y(a,b)$.

For $f(x,y)$, a function of two variables:

$$f_x(a,b)=\lim_{h\to 0}\frac{f(a+h,b)-f(a,b)}{h},\qquad f_y(a,b)=\lim_{h\to 0}\frac{f(a,b+h)-f(a,b)}{h}$$

(The notes’ second formula has a typo, writing $f_y(a,h)$. It means $f_y(a,b)$.)

What are the names and alternate notations for partial derivatives?

• $f_x$ is the partial derivative of $f$ with respect to $x$, or just the $x$-derivative of $f$.

• Other notations: $\dfrac{\partial f}{\partial x}$, $\dfrac{\partial f}{\partial x}\Big|_{(a,b)}$, $\dfrac{\partial f}{\partial y}$, $\dfrac{\partial f}{\partial y}\Big|_{(a,b)}$.

• If $z=f(x,y)$, also write $\dfrac{\partial z}{\partial x},\ \dfrac{\partial z}{\partial y}$.

How do you compute partial derivatives in practice? Work $f(x,y)=x^2y^5$.

• To compute $f_x$, treat $y$ as a constant and differentiate with respect to $x$. To compute $f_y$, treat $x$ as a constant.

• Example: $f_x=2xy^5,\quad f_y=x^2\cdot 5y^4=5x^2y^4$.

Compute $\dfrac{\partial q}{\partial x}$ and $\dfrac{\partial q}{\partial y}$ for $q(x,y)=\dfrac{y^2}{(1+x^2)^3}$.

• x-derivative (treat $y^2$ as constant, use the chain rule):
  $\frac{\partial q}{\partial x}=y^2\frac{\partial}{\partial x}(1+x^2)^{-3}=y^2(-3)(1+x^2)^{-4}(2x)=-6xy^2(1+x^2)^{-4}$

• y-derivative (treat $\frac{1}{(1+x^2)^3}$ as constant):
  $\frac{\partial q}{\partial y}=\frac{1}{(1+x^2)^3}\frac{\partial}{\partial y}(y^2)=\frac{2y}{(1+x^2)^3}$

Compute $\dfrac{\partial}{\partial y}\ln(xy-2y^2)$.

Use the chain rule for $\ln$, treating $x$ as constant:

$$\frac{\partial}{\partial y}\ln(xy-2y^2)=\frac{1}{xy-2y^2}\cdot\frac{\partial}{\partial y}(xy-2y^2)=\frac{x-4y}{xy-2y^2}$$

What is the geometric meaning of $f_x(a,b)$?

• 1-D review: for $y=f(x)$ and the point $(a,f(a))$ on the graph, $f'(a)$ is the slope of the tangent line to the graph.

• 2-D: for $z=f(x,y)$ and the point $(a,b,f(a,b))$ on the graph, set $y=b$ to get the trace curve $z=f(x,b)$.

• Then $f_x(a,b)$ is the slope of the tangent line to the trace curve.

How are partial derivatives defined for more than two variables? Compute $f_z(0,0,1,1)$ for $f(x,y,z,w)=\dfrac{e^{xz+y}}{z^2+w}$.

• Use a similar limit formula, changing only the variable of interest:
  $\frac{\partial f}{\partial z}(x,y,z,w)=f_z=\lim_{h\to 0}\frac{f(x,y,z+h,w)-f(x,y,z,w)}{h}$

• Example: by the quotient rule,
  $f_z=\frac{1}{(z^2+w)^2}\left((z^2+w)\,x\,e^{xz+y}-e^{xz+y}(2z)\right)=\frac{xz^2+xw-2z}{(z^2+w)^2}\,e^{xz+y}$

• At $(0,0,1,1)$:
  $f_z(0,0,1,1)=\frac{(0+0-2)e^0}{(1^2+1)^2}=\frac{-2}{4}=-\frac12$

What are higher-order partial derivatives, and how are they denoted?

• Second-order partials are derivatives of derivatives. For $f(x,y)$ there are 4:
  $f_{xx}=\frac{\partial}{\partial x}(f_x),\quad f_{yy}=\frac{\partial}{\partial y}(f_y),\quad f_{xy}=\frac{\partial}{\partial y}(f_x),\quad f_{yx}=\frac{\partial}{\partial x}(f_y)$

• Other notations: $f_{xx}=\dfrac{\partial^2 f}{\partial x^2}$, $f_{xy}=\dfrac{\partial^2 f}{\partial y\,\partial x}$, $f_{yx}=\dfrac{\partial^2 f}{\partial x\,\partial y}$, $f_{yy}=\dfrac{\partial^2 f}{\partial y^2}$.

• In subscript notation the first letter is the first derivative taken. In $\frac{\partial^2 f}{\partial y\,\partial x}$ the order reads right to left.

• Order 3: there are 8 partials, e.g. $f_{xyy}=\dfrac{\partial}{\partial y}(f_{xy})$.

Compute all first- and second-order partials of $f(x,y)=x^3+y^2e^x$.

• First order: $f_x=3x^2+y^2e^x,\quad f_y=2ye^x$

• Second order:

  • $$f_{xx}=\frac{\partial}{\partial x}(f_x)=6x+y^2e^x$$

  • $$f_{xy}=\frac{\partial}{\partial y}(f_x)=2ye^x$$

  • $$f_{yx}=\frac{\partial}{\partial x}(f_y)=2ye^x$$

  • $$f_{yy}=\frac{\partial}{\partial y}(f_y)=2e^x$$

• Note $f_{xy}=f_{yx}$. This is not a coincidence (see Clairaut’s Theorem).

State Clairaut’s Theorem and its extension to higher orders.

• Theorem: if $f_{xy}$ and $f_{yx}$ both exist and are continuous on a disk $D$, then $f_{xy}(a,b)=f_{yx}(a,b)$ for all $(a,b)$ in $D$. Equivalently,
  $\frac{\partial^2 f}{\partial x\,\partial y}=\frac{\partial^2 f}{\partial y\,\partial x}\quad(\text{or } f_{xy}=f_{yx}).$

• Higher orders: partial derivatives can be computed in any order, e.g.
  $f_{xyxy}=f_{xxyy}=f_{yyxx}=f_{yxyx}=f_{xyyx}=f_{yxxy}.$

• Pitfall: the equality requires the continuity condition.

Compute $g_{zzwx}$ for $g(x,y,z,w)=x^3w^2z^2+\sin\!\left(\dfrac{xy}{z^2}\right)$. What strategy makes it easier?

• Strategy: by Clairaut, reorder the derivatives. Note $\sin\!\left(\frac{xy}{z^2}\right)$ does not depend on $w$, so take the $w$-derivative first to kill it.

• Steps:

  • $$\frac{\partial}{\partial w}\sin\!\left(\frac{xy}{z^2}\right)=0$$

  • $$g_w=2x^3wz^2$$

  • $$g_{wx}=2(3x^2)wz^2=6x^2wz^2$$

  • $$g_{wxz}=6x^2w(2z)=12x^2wz$$

  • $$g_{wxzz}=12x^2w$$

• Therefore $g_{zzwx}=g_{wxzz}=12x^2w$.

What is a PDE, and what does this course expect you to do with one?

• A Partial Differential Equation (PDE) is a differential equation involving multivariable functions and their partial derivatives.

• In this lecture: nothing about solving PDEs. Once a solution is given, you need to know how to check the solution.

• Example (Heat equation): for $u(x,t)$, a function of two variables,
  $\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}$ is a PDE.

Show that $u(x,t)=\dfrac{1}{2\sqrt{\pi t}}\,e^{-x^2/(4t)}$ (for $t>0$) satisfies the heat equation.

Compute both sides and compare.

• $x$-derivatives:
  $\frac{\partial u}{\partial x}=\frac{1}{2\sqrt{\pi t}}\left(-\frac{2x}{4t}\right)e^{-x^2/4t}=-\frac{1}{4\sqrt\pi}\,x\,t^{-3/2}e^{-x^2/4t}$ $\frac{\partial^2 u}{\partial x^2}=-\frac{1}{4\sqrt\pi}t^{-3/2}\left(e^{-x^2/4t}+x\left(-\frac{2x}{4t}\right)e^{-x^2/4t}\right)=-\frac{1}{4\sqrt\pi}t^{-3/2}e^{-x^2/4t}+\frac{1}{8\sqrt\pi}x^2t^{-5/2}e^{-x^2/4t}$

• $t$-derivative (product rule on $t^{-1/2}$ and the exponential):
  $\frac{\partial u}{\partial t}=\frac{1}{2\sqrt\pi}\left(-\frac12 t^{-3/2}e^{-x^2/4t}+t^{-1/2}e^{-x^2/4t}\left(-\frac{x^2}{4}\right)\left(-\frac{1}{t^2}\right)\right)=-\frac{1}{4\sqrt\pi}t^{-3/2}e^{-x^2/4t}+\frac{1}{8\sqrt\pi}t^{-5/2}x^2e^{-x^2/4t}$

• The two expressions match, so $\dfrac{\partial u}{\partial t}=\dfrac{\partial^2 u}{\partial x^2}$.