Lecture 22
3.2.1 Slope in the $x$-direction
Consider the surface $z=f(x,y)=1-x^2-y^2$ and the point $P=(1,-1,-1)$ on the surface. Use the "$y$-is-constant" cross-section through $P$ to find the slope in the $x$-direction at $P.$
3.2.1 Slope in the $x$-direction
Here we have $\,f(x,-1) = 1-x^2-1^2$ $=-x^2$ $=h(x).$
Then $\,h'(x) = -2x.\,$ So $\, h'(1)=-2.$
$\quad -2 $ $=h'(1)$ $=\ds \lim_{\Delta x \to 0} \frac{h(1+\Delta x) - h(1) }{\Delta x}$
$\qquad =\ds \lim_{\Delta x \to 0} \frac{f(1+\Delta x, -1) - f(1,-1) }{\Delta x}$
$\qquad =\ds \frac{\partial f}{\partial x} (1,-1)$
3.2.1 Slope in the $x$-direction
The slope in the $x$-direction, with $y$ held fixed, is called the partial derivative of $f$ with respect to $x$ at the point $(a,b)$
$\ds\frac{\partial f}{\partial x}(a,b) = f_x(a,b) =\lim_{h \ra 0}\frac{f(a+h,b)-f(a,b)}{h} .$
3.2.2 Slope in the $y$-direction
Use the "$x$-is-constant" cross-section to find the slope at $P=(1,-1,-1)$ in the $y$ direction, i.e., where $x=1.$
3.2.2 Slope in the $y$-direction
When $x=1$ we have $$z = f(1,y) = h(y),$$
where $h(y) = -y^2$.
So $h'(y) = -2y\,$ and $\,h'(-1) = -2.$
3.2.2 Slope in the $y$-direction
Similarly, the slope in the $y$-direction, with $x$ held fixed, is called the partial derivative of $f$ with respect to $y$ at the point $(a,b)$
$\ds\frac{\partial f}{\partial y}(a,b) = f_y(a,b) =\lim_{h \ra 0}\frac{f(a,b+h)-f(a,b)}{h} .$
Important remark: Normal rules of differentiation apply, we simply think of the variables being held fixed as constants when doing the differentiation.
3.2.2 Slope in the $y$-direction
Find the partial derivatives $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}\,$ of $f(x,y) = x \sin y + y \cos x.$
3.2.2 Slope in the $y$-direction
Given $f(x,y)=xy^3+x^2,$ find $f_x(1,2)$ and $f_y(1,2).$
3.2.3 Partial derivatives for $f(x,y,z)$
Example: The volume of a box $V(x,y,z)=xyz.$
If $x$ changes by a small amount, say $\Delta x$, denote the corresponding change in $V$ by $\Delta V.$ We can easily visualise that $\Delta V = yz \Delta x.$
3.2.3 Partial derivatives for $f(x,y,z)$
Therefore, $$\dfrac{\Delta V}{\Delta x} = yz.$$
Letting $\Delta x \to 0$ we have $\dfrac{\partial V}{\partial x} = yz$.
For partial derivatives only one independent variable changes and all other independent variables remain fixed.
3.2.3 Partial derivatives for $f(x,y,z)$
Example: Parallel resistance
In an electrical circuit, the combined resistance $R$, from three resistors $R_1, R_2$ and $R_3$ connected in parallel, is \[ \frac{1}{R} = \frac{1}{R_1}+ \frac{1}{R_2} + \frac{1}{R_3}. \] What is the rate of change of the total resistance $R$ with respect to $R_1$?
Example: Parallel resistance $ \quad \ds\frac{1}{R} = \frac{1}{R_1}+ \frac{1}{R_2} + \frac{1}{R_3}. $
3.2.4 Higher order derivatives
The second order partial derivatives of $f$, if they exist, are written as \begin{align*} f_{xx} & = \frac{\partial^2f}{\partial x^2}, & \hspace{-1cm} f_{yx} & = \frac{\partial^2f}{\partial x\partial y}= \frac{\partial}{\partial x}\Bigl(\frac{\partial f}{\partial y}\Bigr), \\[3mm] f_{yy} & = \dfrac{\partial^2f}{\partial y^2}, & \hspace{-1cm} f_{xy} & = \dfrac{\partial^2f}{\partial y\partial x} = \frac{\partial}{\partial y}\Bigl(\frac{\partial f}{\partial x}\Bigr). \end{align*}
If $f_{xy}$ and $f_{yx}$ are both continuous, then $f_{xy} = f_{yx}$.
3.2.4 Higher order derivatives
Example: Returning to the example on page ?? for which $f(x,y) = x \sin y + y \cos x,$ calculate all of the second order partial derivatives of $f$ and show that $$\displaystyle\frac{\partial^2f}{\partial x\partial y}= \dfrac{\partial^2f}{\partial y\partial x}.$$
3.2.4 Higher order derivatives
Example: $f(x,y) = x \sin y + y \cos x,$ calculate all of the second order partial derivatives of $f$ and show that $\displaystyle\frac{\partial^2f}{\partial x\partial y}= \dfrac{\partial^2f}{\partial y\partial x}$.