Calculus 1 - Function of Two or More Variables

Contents: Limit of a Function Derivative of a Function Applications of Derivative Function of Two or More Variables Definition Partial Derivative Extrema Gradient Sample Problems

Definition

Function of Two or More Variables is a rule that assigns each ordered pair (x,y) in the domain of the function a unique real number z. The variable z is usually written as f(x,y). The variables x and y are Independent Variables, and z is a Dependent Variable.

Let f be a function of two variables with the domain D. The graph of f is the set: S={ (x,y,z) | z=f(x,y) , (x,y) ∈ D}
Level curves of a function f of two variables are the curves in the xy-plane with equations f(x,y)=k, where k is a constant in the range of f.

Level curves with the equation f(x,y)=k is the set of all points in the domain of f corresponding to the points on the surface z=f(x,y) having the same height or depth k. Through drawing the level curves corresponding to several admissible values of k, a contour map is obtained.

Limit
Let f be a function that is defined for all points (x,y) close to the point (a,b) with the possible exception of (a,b) itself. Then $\lim_{(x,y)\rightarrow (a,b)} f(x,y)=L$, if f(x,y) can be made as close to L as we please by restricting (x,y) to be sufficiently close to (a,b).

Continuity
Let f be a function that is defined for all points (x,y) close to the point (a,b). Then f is continuous at (a,b) if:
$\lim_{(x,y)\rightarrow (a,b)} f(x,y)=f(a,b)$

Partial Derivative

Let us define a function f, where z=f(x,y):
The partial derivative (First order) of the function f with respect to x:
$z_x=\frac{\partial f}{\partial x} =\lim_{h\rightarrow 0} \frac{f(x+h,y)-f(x,y)}{h}$

The partial derivative (First order) of the function f with respect to y:
$z_y=\frac{\partial f}{\partial y} =\lim_{h\rightarrow 0} \frac{f(x,y+h)-f(x,y)}{h}$

The partial derivative (Second order) of the function f:

$\begin{align*} z_{xx}&=\frac{\partial ^2 z}{\partial x^2} = \frac{\partial }{\partial x} \left (\frac{\partial z}{\partial x} \right )\\ z_{xy}&=\frac{\partial ^2 z}{\partial y \partial x} = \frac{\partial }{\partial y} \left (\frac{\partial z}{\partial x} \right )\\ z_{yx}&=\frac{\partial ^2 z}{\partial x \partial y} = \frac{\partial }{\partial x} \left (\frac{\partial z}{\partial y} \right )\\ z_{yy}&=\frac{\partial ^2 z}{\partial y^2} = \frac{\partial }{\partial y} \left (\frac{\partial z}{\partial y} \right )\\ \end{align*}$

The total derivative of f:
$dz=\frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y}dy$

Partial Derivative of Implicit Function
If f(x,y)=0 then:
$\frac{dy}{dx}= - \frac{f_x}{f_y}$

If f(x,y,z)=0 then:
$\begin{align*} z_x &= \frac{\partial z}{\partial x}= - \frac{f_x}{f_z}\\ z_y &= \frac{\partial z}{\partial y}= - \frac{f_y}{f_z} \end{align*}$

Chain Rule (One Independent Variable)
Let z=f(x,y), where f is differentiable function of x and y. If x=g(t) and y=h(t), where g and h are differentiable functions of t, then z is a differentiable function of t:
$\begin{align*} \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} \end{align*}$

Chain Rule (Two Independent Variable)
Let z=f(x,y), where f is differentiable function of x and y. If x=g(u,v) and y=h(u,v), where g and h are differentiable functions of u and v:
$\begin{align*} \frac{\partial z}{\partial u} &= \frac{\partial z}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial u}\\ \frac{\partial z}{\partial v} &= \frac{\partial z}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial v} \end{align*}$

Extrema

Relative Extrema
Let f be a function defined on a region R containing the point (a,b). Then, the relative maximum of f is at (a,b) if (x,y) ≤ f(a,b) for all points (x,y) in an open disk containing (a,b). The number f(a,b) is called a relative maximum value.

Likewise, the relative minimum of f is at (a,b) if (x,y) ≥ f(a,b) for all points (x,y) in an open disk containing (a,b). The number f(a,b) is called a relative minimum value.

Absolute Extrema
If the relative maximum / minimum at (a,b) holds for all points (x,y), then that (a,b) is the absolute maximum / minimum.

Critical Points
Let f be defined on an open region R containing the point (a,b). (a,b) is the critical point of f if:

• f_x and/or f_y don't exist at (a,b), or
• both f_x(a,b)=0 and f_y(a,b)=0.

If f has a relative extremum at the point (a,b), then (a,b) is a critical point of f.

Second Derivative Test Suppose that f has continuous second-order partial derivatives on an open region containing a critical point (a,b) of f. Let:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-f^{2}_{xy}(x,y)$

• If D(a,b) > 0, and f_xx(a,b) < 0, then f(a,b) is a relative maximum value.
• If D(a,b) > 0, and f_xx(a,b) > 0, then f(a,b) is a relative minimum value.
• If D(a,b) < 0, then (a, b, f(a,b)) is a saddle point.

Gradient

Suppose that f is function of two variables x and y. Its gradient is the vector function:
$\nabla f(x,y) = f_x \overrightarrow{i} + f_y \overrightarrow{j}$

If f(x,y) = 0 and (a,b) is on the curve, then:

• Tangent Line: f_x(x-a) + f_y(y-b) = 0
• Normal Line: (x,y) = (a,b) + λ(f_x,f_y)

Suppose that f is a function of three variables x y and z, then its gradient is the vector function:
$\nabla f(x,y,z) = f_x \overrightarrow{i} + f_y \overrightarrow{j} + f_z \overrightarrow{k}$

If f(x,y,z) = 0 and (a,b,c) is on the curve, then:

• Tangent Line: f_x(x-a) + f_y(y-b) + f_z(z-c) = 0
• Normal Line: (x,y,z) = (a,b,c) + λ(f_x,f_y,f_z)

Sample Problems

Problem 1: Identifying Domain
Determine the domain of the given functions:

$\begin{align*} \text{A. } &f(x,y)=\sqrt{4-x^2-y^2}\\ \text{B. } &f(x,y)=\ln{(x+y)}\\ \text{C. } &f(x,y)=\frac{\sqrt{x-y}}{x-1}\\ \text{D. } &f(x,y,z)=\cos^{-1}(x+y+z)\\ \text{E. } &f(x,y,z)=\frac{\sqrt{4-x^2-4y^2}}{x+y-z}\\ \end{align*}$

Keep in mind that logarithms can't be negative and zero, square roots can't be negative, and denominator can't be zero.

Sub-Problem A
Square roots cannot be negative. So x²+y² must be less than or equal to 4.
The domain of f is {(x,y) ∈ R | x²+y² ≤ 4}

Sub-Problem B
Logarithm cannot be negative, or zero. So, x+y must be greater than 4.
The domain of f is {(x,y) ∈ R | x + y > 0}

Sub-Problem C
Square roots cannot be negative. So, x must be greater than or equal to y. The denominator cannot be zero, which means x must not be equal to 1.
The domain of f is {(x,y) ∈ R | x ≥ y, x ≠ 1}

Sub-Problem D
The value of cosine ranges from -1 to 1. So,
The domain of f is {(x,y,z) ∈ R | 1 ≥ |x+y+z|}

Sub-Problem E
Square roots cannot be negative. So, x²+4y² must be less than or equal to 4. Denominator cannot be zero. So,
The domain of f is {(x,y,z) ∈ R | x + y ≠ z, x²+4y² ≤ 4}

Problem 2: Finding Derivative
Find all of the first and second order partial derivatives, and the total derivative of the following function:

$\begin{align*} \text{A. } &f(x,y)=x^2+xy+y^2\\ \text{B. } &f(x,y)=x^3y^2 + \ln{(x^2)} - \cot{(xy)}\\ \text{C. } &f(x,y)=\frac{1}{x-y}\\ \text{D. } &f(x,y)=e^{xy}-4\sin{xy}+x^4y\\ \text{E. } &f(x,y)=\sin{\left ( \frac{y}{x} \right )} + y^5x^4 -x^y \ln{(x)} \\ \end{align*}$

In partial differentiation, it is pretty much similiar to the single variable function in the previous topics. Which means that, we can use the derivation rules we know. When derivating a function with respect to a variable (x for example), other variables in that functions are treated constants.

Sub-Problem A
$\begin{align*} z&=x^2+xy+y^2\\ \\ z_{x}&=2x+y\\ z_{y}&=2y+x\\ \\ z_{xx}&=2\\ z_{xy}&=1\\ \\ z_{yx}&=1\\ z_{yy}&=2\\ \\ dz&=(2x+y) dx + (2y+x) dy \end{align*}$

Sub-Problem B
$\begin{align*} z&=x^3y^2 + \ln{(x^2)} - \cot{(xy)}\\ &=x^3y^2 + 2\ln{(x)} - \cot{(xy)}\\ \\ z_{x}&=3x^2y^2+\frac{2}{x}+y\csc^2{(xy)}\\ z_{y}&=2x^3y + x \csc^2{(xy)}\\ \\ z_{xx}&=6xy^2-\frac{2}{x^2}-2y^2\csc^2{(xy)}\cot{(xy)} \\ z_{xy}&=6x^2y+csc^2{(xy)}-2xy\csc^2{(xy)}\cot{(xy)}\\ \\ z_{yx}&=6x^2y + \csc^2{(xy)} - 2xy\csc^2{(xy)}\cot{(xy)}\\ z_{yy}&=2x^3 - 2x^2 \csc^2{(xy)} \cot{(xy)}\\ \\ dz &= \left ( 3x^2y^2+\frac{2}{x}+y\csc^2{(xy)} \right ) dx + \left ( 2x^3y + x \csc^2{(xy)}\right ) dy \end{align*}$

Sub-Problem C
$\begin{align*} z&=\frac{1}{x-y}\\ &=(x-y)^{-1}\\ \\ z_{x}&=-(x-y)^{-2}\\ z_{y}&=(x-y)^{-2}\\ \\ z_{xx}&=2(x-y)^{-3} \\ z_{xy}&=-2(x-y)^{-3} \\ \\ z_{yx}&=-2(x-y)^{-3} \\ z_{yy}&=2(x-y)^{-3} \\ \\ dz&=-\frac{1}{(x-y)^2}dx + \frac{1}{(x-y)^2}dy \end{align*}$

Sub-Problem D
$\begin{align*} z&=e^{x+y}-4\sin{xy}+x^4y\\ &=e^{x} e^{y} - 4\sin{xy}+x^4y\\ \\ z_{x}&=e^{x} e^{y} - 4y\cos{xy}+4x^3y\\ z_{y}&=e^{x} e^{y} - 4x\cos{xy}+x^4\\ \\ z_{xx}&=e^{x} e^{y} + 4y^2\sin{xy}+12x^2y\\ z_{xy}&=e^{x} e^{y} - (4\cos{xy} - 4xy\sin{xy}) + 4x^3y\\ \\ z_{yx}&=e^{x} e^{y} - (4\cos{xy} - 4xy\sin{xy}) +4x^3 \\ z_{yy}&=e^{x} e^{y} + 4x^2\sin{xy}\\ \\ dz&=(e^{x} e^{y} - 4y\cos{xy}+4x^3y) dx + (e^{x} e^{y} - 4x\cos{xy}+x^4) dy \end{align*}$

Sub-Problem E
$\begin{align*} z&=\sin{\left ( \frac{y}{x} \right )} + y^5x^4 - x^y \ln{(x)}\\ \\ z_{x}&=\frac{-y \cos{\left ( \frac{y}{x} \right )}}{x^2} + 4y^5x^3 - yx^{y-1} \ln{(x)} + x^{y-1}\\ z_{y}&=\frac{\cos{\left ( \frac{y}{x} \right )}}{x} + 5y^4x^4 - x^y (\ln{(x)})^2\\ \\ z_{xx}&=\frac{2y\cos{\left ( \frac{y}{x} \right )}}{x^3} - \frac{y^2 \sin{\left ( \frac{y}{x} \right )}}{x^2} + 12x^2y^5 - (y^2-y)x^{y-2} \ln{x} -x^{y-2}\\ z_{xy}&=-\frac{1}{x^2}\cos{\left ( \frac{y}{x} \right )} - \frac{y}{x^3}\sin{\left ( \frac{y}{x} \right ) } + 20y^4x^3 - yx^{y-1} (\ln{(x)})^2\\ z_{yx}&=\frac{y \sin {\left ( \frac{y}{x} \right )} - x \cos{\left ( \frac{y}{x} \right )}}{x^3} + 20y^4x^3 - yx^{y-1} (\ln{(x)})^2 - 2x^{y-1}\ln{(x)}\\ z_{yy}&=-\frac{\sin{\left ( \frac{y}{x} \right )}}{x^2} + 20y^3x^4 - x^y(\ln{(x)})^3\\ \\ dz&=\left (\frac{-y \cos{\left ( \frac{y}{x} \right )}}{x^2} + 4y^5x^3 - yx^{y-1} \ln{(x)} + x^{y-1} \right )dx+ \left ( \frac{\cos{\left ( \frac{y}{x} \right )}}{x} + 5y^4x^4 - x^y (\ln{(x)})^2 \right ) dy\\ \end{align*}$

Problem 3: Finding Derivative
Given a multi-variable function w, find dw/dt using the chain rule:

$\begin{align*} \text{A. } &w=x^2y\;;x=t^2\;;y=t^3\\ \text{B. } &w=\frac{\ln{(x)}}{\ln{(y)}}\;;x=t+1\;;y=t-1\\ \text{C. } &w=\tan{(xy)} + x^5y^4 - \frac{3 x}{y^2}\;;x=\sin^2{(t)}\;;y=\csc{(t)}\\ \text{D. } &w=\tan^{-1}(xz)+\frac{z}{y}\;;x=t\;;y=t^2\;;z=\sinh{t}\\ \text{E. } &w=x \sqrt{y^2+z^2}\;;x=\frac{1}{t}\;;y=e^{-t}\cos{t}\;;z=e^{-t}\sin{t}\\ \end{align*}$

Derivation of w=f(x,y):
$\begin{align*} \frac{dw}{dt}&= \left ( \frac{\partial w}{\partial x} \right )\left ( \frac{dx}{dt} \right ) + \left ( \frac{\partial w}{\partial y} \right )\left ( \frac{dy}{dt} \right ) \\ \end{align*}$

Derivation of w=f(x,y,z):
$\begin{align*} \frac{dw}{dt}&= \left ( \frac{\partial w}{\partial x} \right )\left ( \frac{dx}{dt} \right ) + \left ( \frac{\partial w}{\partial y} \right )\left ( \frac{dy}{dt} \right ) + \left ( \frac{\partial w}{\partial z} \right )\left ( \frac{dz}{dt} \right )\\ \end{align*}$

Sub-Problem A
$\begin{align*} w&=x^2y\\ x&=t^2\\ y&=t^3\\ \\ \frac{\partial w}{\partial x} &= 2xy\\ \frac{\partial w}{\partial y} &= x^2\\ \\ \frac{dx}{dt} &= 2t\\ \frac{dy}{dt} &= 3t^2\\ \\ \frac{dw}{dt} &= (2xy)(2t) + (x^2)(3t^2)\\ &=(2t^5)(2t)+(t^4)(3t^2)\\ &=7t^6 \end{align*}$

Sub-Problem B
$\begin{align*} w&=\frac{\ln{x}}{\ln{y}}\\ \\ x&=t+1\\ y&=t-1\\ \\ \frac{\partial w}{\partial x} &= \frac{1}{x\ln{y}}\\ &= \frac{1}{(t+1)\ln{(t-1)}}\\ \frac{\partial w}{\partial y} &= -\frac{\ln{x}}{y (\ln{y})^2}\\ &= -\frac{\ln{(t+1)}}{(t-1) (\ln{(t-1)})^2 }\\ \\ \frac{dx}{dt} &= 1\\ \frac{dy}{dt} &= 1\\ \\ \frac{dw}{dt} &= \frac{1}{(t+1) \ln{(t-1)}} - \frac{\ln{(t+1)}}{(t-1) (\ln{(t-1)})^2 } \end{align*}$

Sub-Problem C

$\begin{align*} w&=\tan{(xy)} + x^5y^4 - \frac{3 x}{y^2}\\ \\ x&=\sin^2{t}\\ y&=\csc{t}\\ xy&=\sin{t}\\ xy^2&=1\\ \\ \frac{\partial w}{\partial x} &= y \sec^{2}{(xy)} + 5x^4y^4 - \frac{3}{y^2}\\ &= \csc{(t)}\sec^{2}{(\sin{(t)})} + 5(\sin{(x)})^4 - 3\sin^2{(t)}\\ \frac{\partial w}{\partial y} &= x \sec^{2}{(xy)} + 4x^5y^3 + \frac{6x}{y^3}\\ &= \sin^2{(t)} \sec^{2}{(\sin{(t)})} + 4(\sin{(t)})^7 + 9(\sin{(t)})^5\\ \\ \frac{dx}{dt} &= \sin{(2t)}\\ \frac{dy}{dt} &= -\csc{(t)}\cot{(t)}\\ \\ \frac{dw}{dt} &= (\csc{(t)}\sec^{2}{(\sin{(t)})} + 5(\sin{(x)})^4 - 3\sin^2{(t)}) (\sin{(2t)}) + (\sin^2{(t)} \sec^{2}{(\sin{(t)})} + 4(\sin{(t)})^7 + 9(\sin{(t)})^5) (-\csc{(t)}\cot{(t)}) \end{align*}$

Sub-Problem D

$\begin{align*} &w=\tan^{-1}(xz)+\frac{z}{y}\;;x=t\;;y=t^2\;;z=\sinh{(t)}\\ \\ x&=t\\ y&=t^2\\ z&=\sinh{t}\\ \\ \frac{\partial w}{\partial x} &= \frac{z}{(xz)^2+1}\\ &=\frac{\sinh{(t)}}{(t^2 \sinh^2{(t)}) + 1 }\\ \frac{\partial w}{\partial y} &= -\frac{z}{y^2}\\ &= - \frac{\sinh{(t)}}{t^4}\\ \frac{\partial w}{\partial z} &= \frac{x}{(xz)^2+1} + \frac{1}{y}\\ &= \frac{t}{t^2 \sinh^2{(t)} + 1} + \frac{1}{t^2}\\ \\ \frac{dx}{dt} &= 1\\ \frac{dy}{dt} &= 2t\\ \frac{dz}{dt} &= \cosh{(t)}\\ \\ \frac{dw}{dt} &= \frac{\sinh{(t)}}{t^2 \sinh^2{(t)} + 1 } - \frac{2\sinh{(t)}}{t^3} + \frac{t\cosh{(t)}}{t^2 \sinh^2{(t)} + 1} + \frac{\cosh{(t)}}{t^2}\\ &= \frac{\sinh{(t)} + t\cosh{(t)}}{t^2 \sinh^2{(t)} + 1 } + \frac{t\cosh{(t)} - 2 \sinh{(t)}} {t^3}\\ \end{align*}$

Sub-Problem E

$\begin{align*} w&=x \sqrt{y^2+z^2}\\ \\ x&=\frac{1}{t}\\ y&=e^{-t}\cos{t}\\ z&=e^{-t}\sin{t}\\ \\ \frac{\partial w}{\partial x} &= \sqrt{y^2+z^2}\\ &= \sqrt{e^{-2t}\cos^2{t}+e^{-2t}\sin^2{t}}\\ &=e^{-t}\\ \frac{\partial w}{\partial y} &= \frac{xy}{\sqrt{y^2+z^2}}\\ &=\frac{xy}{e^{-t}}\\ &=\frac{\cos{t}}{t}\\ \frac{\partial w}{\partial z} &= \frac{xz}{\sqrt{y^2+z^2}}\\ &=\frac{xz}{e^{-t}}\\ &=\frac{\sin{t}}{t}\\ \\ \frac{dx}{dt} &= -\frac{1}{t^2}\\ \frac{dy}{dt} &= -e^{-t}\cos(t) - e^{-t}\sin(t) \\ &=-e^{-t} (\sin{t}+\cos{t})\\ \frac{dz}{dt} &= -e^{-t}\sin(t) + e^{-t}\cos(t) \\ &=e^{-t} (\cos{t}-\sin{t})\\ \\ \frac{dw}{dt} &= e^{-t}\left (-\frac{1}{t^2} \right ) + \frac{\cos{t}}{t} \left ( -e^{-t} (\sin{t}+\cos{t}) \right ) + \frac{\sin{t}}{t}(e^{-t} (\cos{t}-\sin{t}))\\ &= \frac{-e^{-t} + t \cos{t}\left ( -e^{-t} (\sin{t}+\cos{t}) \right ) + t \sin{t} (e^{-t} (\cos{t}-\sin{t}))}{t^2} \end{align*}$

Problem 4: Finding Derivative
Given a multi-variable function w, find ∂w/∂s and ∂w/∂t using the chain rule:

$\begin{align*} \text{A. } w&=e^x \cos{y}\;;x=\ln(s^2+t^2)\;;y=\sqrt{st}\\ \text{B. } w&=x^3+y^3\;;x=s^2+t^2\;;y=2st\\ \text{C. } w&=x \csc{yz}\;;x=rs\;;y=s^2t\;; z = \frac{s}{t^2}\\ \text{D. } w&=\frac{x+y}{x+z}\;;x=r\cos{s}\;;y=r \sin{t}\;; z = s \tan{t}\\ \text{E. } w&=x \tan^{-1}(yz)\;;x=\sqrt{s}\;;y=e^{-2t}\;;z=t \cos{s}\\ \\ \end{align*}$

Partial derivation of w=f(x,y):
$\begin{align*} \frac{\partial w}{\partial s}&= \left ( \frac{\partial w}{\partial x} \right )\left ( \frac{\partial x}{\partial s} \right ) + \left ( \frac{\partial w}{\partial y} \right )\left ( \frac{\partial y}{\partial s} \right ) \\ \frac{\partial w}{\partial t}&= \left ( \frac{\partial w}{\partial x} \right )\left ( \frac{\partial x}{\partial t} \right ) + \left ( \frac{\partial w}{\partial y} \right )\left ( \frac{\partial y}{\partial t} \right ) \\ \end{align*}$

Partial derivation of w=f(x,y,z):
$\begin{align*} \frac{\partial w}{\partial s}&= \left ( \frac{\partial w}{\partial x} \right )\left ( \frac{\partial x}{\partial s} \right ) + \left ( \frac{\partial w}{\partial y} \right )\left ( \frac{\partial y}{\partial s} \right ) + \left ( \frac{\partial w}{\partial z} \right )\left ( \frac{\partial z}{\partial s} \right )\\ \frac{\partial w}{\partial t}&= \left ( \frac{\partial w}{\partial x} \right )\left ( \frac{\partial x}{\partial t} \right ) + \left ( \frac{\partial w}{\partial y} \right )\left ( \frac{\partial y}{\partial t} \right )+ \left ( \frac{\partial w}{\partial z} \right )\left ( \frac{\partial z}{\partial t} \right ) \\ \end{align*}$

Sub-Problem A

$\begin{align*} w&=e^x \cos{y}\\ \\ x&=\ln{(s^2+t^2)}\\ y&=\sqrt{st}\\ \\ \frac{\partial w}{\partial x}&= e^x \cos{y}\\ &= e^{\ln (s^2+t^2)} \cos{\sqrt{st}}\\ &= (s^2+t^2) (\cos{\sqrt{st}})\\ \frac{\partial w}{\partial y}&= -e^x \sin{y}\\ &= -(s^2+t^2) (\sin{\sqrt{st}})\\ \\ \frac{\partial x}{\partial s}&= \frac{2s}{s^2+t^2}\\ \frac{\partial y}{\partial s}&= \frac{1}{2}\sqrt{\frac{t}{s}}\\ \\ \frac{\partial x}{\partial t}&= \frac{2t}{s^2+t^2}\\ \frac{\partial y}{\partial t}&= \frac{1}{2}\sqrt{\frac{s}{t}}\\ \\ \frac{\partial w}{\partial s}&= \left ( \frac{\partial w}{\partial x} \right )\left ( \frac{\partial x}{\partial s} \right ) + \left ( \frac{\partial w}{\partial x} \right )\left ( \frac{\partial x}{\partial s} \right ) \\ &=2s(\cos{\sqrt{st}})-\frac{\sqrt{t}(s^2+t^2)(\sin{\sqrt{st}})}{2\sqrt{s}}\\ \frac{\partial w}{\partial t}&= \left ( \frac{\partial w}{\partial x} \right )\left ( \frac{\partial x}{\partial t} \right ) + \left ( \frac{\partial w}{\partial x} \right )\left ( \frac{\partial x}{\partial t} \right ) \\ &=2t(\cos{\sqrt{st}})-\frac{\sqrt{s}(s^2+t^2)(\sin{\sqrt{st}})}{2\sqrt{t}}\\ \end{align*}$

Sub-Problem B

$\begin{align*} w&=x^3+y^3\\ \\ x&=s^2+t^2\\ y&=2st\\ \\ \frac{\partial w}{\partial x} &= 3x^2\\ \frac{\partial w}{\partial y} &= 3y^2\\ \\ \frac{\partial x}{\partial s} &= 2s\\ \frac{\partial y}{\partial s} &= 2t\\ \\ \frac{\partial x}{\partial t} &= 2t\\ \frac{\partial y}{\partial t} &= 2s\\ \\ \frac{\partial w}{\partial s} &= \frac{\partial w}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial s} \\ &=(3x^2)(2s)+(3y^2)(2t)\\ &=6(x^2s+y^2t)\\ &=6((s^2+t^2)^2s + (2st)^2t) \\ \frac{\partial w}{\partial t} &= \frac{\partial w}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial t} \\ &= (3x^2)(2t) + (3y^2)(2s)\\ &=6(x^2t+y^2s)\\ &=6((s^2+t^2)^2t + (2st)^2s)\\ \\ \end{align*}$

Sub-Problem C

$\begin{align*} w&=x \csc{yz}\\ \\ x&=rs\\ y&=s^2t\\ z&=\frac{s}{t^2}\\ \\ \frac{\partial w}{\partial x} &= \csc{yz}\\ &=\csc \left ( \frac{s^3}{t} \right )\\ \frac{\partial w}{\partial y} &= -xz \csc{(yz)} \cot{(yz)}\\ &= -\frac{rs^2}{t^2} \csc \left ( \frac{s^3}{t} \right ) \cot \left ( \frac{s^3}{t} \right )\\ \frac{\partial w}{\partial z} &= -xy \csc{(yz)} \cot{(yz)}\\ &=-rs^3t\csc \left ( \frac{s^3}{t} \right ) \cot \left ( \frac{s^3}{t} \right )\\ \\ \frac{\partial x}{\partial s} &= r\\ \frac{\partial y}{\partial s} &= 2st\\ \frac{\partial z}{\partial s} &= \frac{1}{t^2}\\ \\ \frac{\partial x}{\partial t} &= 0\\ \frac{\partial y}{\partial t} &= s^2\\ \frac{\partial z}{\partial t} &= -\frac{2s}{t^3}\\ \\ \frac{\partial w}{\partial s}&= r \csc{\left ( \frac{s^3}{t}\right )} - \frac{2rs^3}{t} \csc \left ( \frac{s^3}{t} \right ) \cot \left ( \frac{s^3}{t} \right ) -\frac{rs^3}{t}\csc \left ( \frac{s^3}{t} \right ) \cot \left ( \frac{s^3}{t} \right ) \\ &= r \csc{\left ( \frac{s^3}{t}\right )} - \frac{3rs^3}{t} \csc \left ( \frac{s^3}{t} \right ) \cot \left ( \frac{s^3}{t} \right )\\ \frac{\partial w}{\partial t}&= -\frac{rs^4}{t^2} \csc \left ( \frac{s^3}{t} \right ) \cot \left ( \frac{s^3}{t} \right ) +\frac{2rs^4}{t^2}\csc \left ( \frac{s^3}{t} \right ) \cot \left ( \frac{s^3}{t} \right )\\ &= \frac{rs^4}{t^2} \csc \left ( \frac{s^3}{t} \right ) \cot \left ( \frac{s^3}{t} \right ) \end{align*}$

Sub-Problem D

$\begin{align*} w&=\frac{x+y}{x+z}\\ \\ x&=r \cos{s}\\ y&=r \sin{t}\\ z&=s \tan{t}\\ \\ \frac{\partial w}{\partial x} &= \frac{(x+z)-(x+y)}{(x+z)^2}\\ &= \frac{z-y}{(x+z)^2}\\ \frac{\partial w}{\partial y} &= \frac{1}{x+z}\\ \frac{\partial w}{\partial z} &= \frac{-x-y}{(x+z)^2}\\ \\ \frac{\partial x}{\partial s} &= -r \sin{s}\\ \frac{\partial y}{\partial s} &= 0\\ \frac{\partial z}{\partial s} &= \tan{t}\\ \\ \frac{\partial x}{\partial t} &= 0\\ \frac{\partial y}{\partial t} &= r\cos{t}\\ \frac{\partial z}{\partial t} &= s \sec^2{t}\\ \\ \frac{\partial w}{\partial s}&= (-r \sin{s})\frac{z-y}{(x+z)^2} - (\tan{t}) \frac{x+y}{(x+z)^2}\\ &= \frac{-r z \sin{s} +r y \sin{s} +x\tan{t}+y\tan{t}}{(x+z)^2}\\ &= \frac{-rs\sin{s}\tan{t} + r^2 \sin{s}\sin{t} - r \cos{s} \tan{t} - r \tan{t} \sin{t}}{(r \cos{s}+s \tan{t})^2}\\ \frac{\partial w}{\partial t}&= ( r \cos{t})\frac{1}{x+z} - (s \sec^2{t}) \frac{x+y}{(x+z)^2}\\ &= \frac{(r \cos{t})(r \cos{s}+s \tan{t}) - (s \sec^2{t})(r\cos{t} + r \sin{t})}{(r \cos{s}+s \tan{t})^2}\\ \end{align*}$

Sub-Problem E

$\begin{align*} w&=x \tan^{-1}(yz)\\ \\ x&=\sqrt{s}\\ y&=e^{-2t}\\ z&=t \cos{s}\\ \\ \frac{\partial w}{\partial x} &= \tan^{-1}(yz)\\ \frac{\partial w}{\partial y} &= \frac{xz}{(yz)^2+1}\\ \frac{\partial w}{\partial z} &= \frac{xy}{(yz)^2+1}\\ \\ \frac{\partial x}{\partial s} &= \frac{1}{2\sqrt{s}}\\ \frac{\partial y}{\partial s} &= 0\\ \frac{\partial z}{\partial s} &= -t\sin{s}\\ \\ \frac{\partial x}{\partial t} &= 0\\ \frac{\partial y}{\partial t} &= -2e^{-2t}\\ \frac{\partial z}{\partial t} &= \cos{s}\\ \\ \frac{\partial w}{\partial s} &= \frac{\tan^{-1}(yz)}{2\sqrt{s}} - \frac{txy \sin{s}}{(yz)^2+1}\\ &=\frac{\tan^{-1}(e^{-2t} t \cos{s})}{2\sqrt{s}} - \frac{t\sqrt{s} e^{-2t} \sin{s}}{(e^{-2t} t \cos{s})^2+1} \\ \frac{\partial w}{\partial t} &= \frac{xz(-2e^{-2t})}{(yz)^2+1} + \frac{xy \cos{s}}{(yz)^2+1}\\ &= \frac{xz(-2e^{-2t}) + xy \cos{s}}{(yz)^2+1}\\ &= \frac{t \sqrt{s} \cos{(s)} (-2e^{-2t}) + \sqrt{s}e^{-2t} \cos{s}}{(e^{-2t} t \cos{s})^2+1}\\ &= \frac{(e^{-2t}\sqrt{s} \cos{s}) (-2t+1)}{(e^{-2t} t \cos{s})^2+1} \end{align*}$

Problem 5: Finding Derivative
A-C: Given a function f(x,y)=0, find dy/dx through the formula:
$\frac{dy}{dx}= - \frac{f_x}{f_y}$.
D-E: Given a function f(x,y,z)=0, find ∂z/∂x and ∂z/∂y through the following formulas:
$\begin{align*} \frac{\partial z}{\partial x}= - \frac{f_x}{f_z}\\ \frac{\partial z}{\partial y}= - \frac{f_y}{f_z} \end{align*}$.

$\begin{align*} \text{A. } &x^2+3y^2-4x+3y+2=0\\ \text{B. } &x^3-2xy+y^3=4\\ \text{C. } &2x^2+3\sqrt{xy}-2y=4\\ \text{D. } &x^2+y^2+z^2=16\\ \text{E. } &x^2+y^3-z+2yz^2=5\\ \end{align*}$

After we've used the formula above, we can later check if our answers are correct or not by performing the regular implicit differentiation (which has been done in previous topic).

Sub-Problem A
$\begin{align*} &x^2+3y^2-4x+3y+2=0\\ \\ f_x&=2x-4\\ f_y&=6y+3\\ \\ \frac{dy}{dx} &= -\frac{2x-4}{6y-3} \\ \end{align*}$

Sub-Problem B
$\begin{align*} &x^3-2xy+y^3=4\\ \\ &x^3-2xy+y^3-4=0\\ \\ f_x&=3x^2-2y\\ f_y&=-2x+3y^2\\ \\ \frac{dy}{dx} &= -\frac{3x^2-2y}{3y^2-2x} \\ \end{align*}$

Sub-Problem C
$\begin{align*} &2x^2+3\sqrt{xy}-2y=4\\ \\ &2x^2+3\sqrt{xy}-2y-4=0\\ \\ f_x&=4x + \frac{3\sqrt{y}}{\sqrt{x}}\\ f_y&=-2 + \frac{3\sqrt{x}}{\sqrt{y}}\\ \\ \frac{dy}{dx} &= -\frac{4x+\frac{3\sqrt{y}}{\sqrt{x}}}{-2+3\frac{\sqrt{x}}{\sqrt{y}}} \\&= -\frac{4x\sqrt{xy}+3y}{-2\sqrt{xy}+3x} \\ \end{align*}$

Sub-Problem D
$\begin{align*} &x^2+y^2+z^2=16\\ \\ &x^2+y^2+z^2-16=0\\ \\ f_x&= 2x\\ f_y&= 2y\\ f_z&= 2z\\ \\ \frac{\partial z}{\partial x} &= - \frac{x}{z}\\ \frac{\partial z}{\partial y} &= - \frac{y}{z}\\ \end{align*}$

Sub-Problem E
$\begin{align*} &x^2+y^3-z+2yz^2=5\\ \\ &x^2+y^3-z+2yz^2-5=0\\ \\ f_x&= 2x\\ f_y&= 3y^2+2z^2\\ f_z&= 4yz-1\\ \\ \frac{\partial z}{\partial x} &= - \frac{2x}{4yz-1}\\ \frac{\partial z}{\partial y} &= - \frac{3y^2+2z^2}{4zy-1}\\ \end{align*}$

Problem 6: Finding Extrema
Given a function f, determine the relative extrema and saddle points.

$\begin{align*} \text{A. } &f(x,y)=x^2+y^2-2x+4y\\ \text{B. } &f(x,y)=e^{-x^2-y^2}\\ \text{C. } &f(x,y)=x\sin{y}\;\;,x\geq0\;\;,0\leq y\leq \pi\\ \text{D. } &f(x,y)=e^{-x} \cos{y}\;\;,x\geq0\;\;,0\leq y\leq \pi\\ \text{E. } &f(x,y)=\frac{1}{2}x^4 - 2x^3 + 4xy + y^2\\ \end{align*}$

Sub-Problem A
$\begin{align*} f(x,y)&=x^2+y^2-2x+4y\\ \\ f_x&=2x-2\\ 0&=2x-2\\ x&=1\\ \\ f_y&=2y+4\\ 0&=2y+4\\ y&=-2\\ \\ f_{xx} &= 2\\ f_{xy} &= 0\\ f_{yy} &= 2\\ \\ D(1,-2) &= f_{xx}(1,-2) f_{yy}(1,-2) - (f_{xy}(1,-2))^2\\ &=2(2) - 0^2\\ &=4\\ \end{align*}$

Since f_xx(1,-2) is greater than 0 and D(1,-2) > 0, it can be concluded that relative minimum occurs during x=1 and y=-2 where f(1,-2) = -5.

Sub-Problem B
$\begin{align*} f(x,y)&=e^{-x^2-y^2}\\ \\ f_x&=e^{-x^2-y^2} (-2x)\\ 0&=e^{-x^2-y^2} (-2x)\\ x&=0\\ \\ f_y&=e^{-x^2-y^2} (-2y)\\ y&=0\\ \\ f_{xx} &= e^{-x^2-y^2} (-2x)^2 -2 e^{-x^2-y^2}\\ &= e^{-x^2-y^2} (-4x^2-1)\\ f_{xy} &= 0\\ f_{yy} &= e^{-x^2-y^2} (-4y^2-1)\\ \\ D(0,0) &= f_{xx}(0,0) f_{yy}(0,0) - (f_{xy}(0,0))^2\\ &=(-1)(-1)\\ &=1\\ \end{align*}$

Since f_xx(0,0) is less than 0 and D(0,0) > 0, it can be concluded that relative maximum occurs during x=0 and y=0 where f(0,0) = 1.

Sub-Problem C
$\begin{align*} f(x,y)&=x \sin{y}\\ \\ f_x&=\sin{y}\\ 0&=\sin{y}\\ y&=m \pi,\text{where }m \in \mathbb{Z} \\ \\ f_y&=xy \cos{y}\\ x&=0\\ \\ f_{xx} &= 0\\ f_{xy} &= \cos{y}\\ f_{yy} &= x \cos{y} - xy\sin{y}\\ \\ D(0,m \pi) &= f_{xx}(0,m \pi) f_{yy}(0,m \pi) - (f_{xy}(0,m \pi))^2\\ D(0,0) &= 0 - 1\\ &=-1\\ D(0,\pi) &= 0 - 1\\ &=-1\\ D(0,2\pi) &= 0 - 1\\ &=-1\\ \end{align*}$

We have three critical points:

• Since D(0,0) < 0, it can be concluded that saddle point occurs during x=0 and y=0. The saddle point will be (0,0,0).
• Since D(0,π) < 0, it can be concluded that saddle point occurs during x=0 and y=π. The saddle point will be (0,π,0).
• Since D(0,2π) < 0, it can be concluded that saddle point occurs during x=0 and y=2π. The saddle point will be (0,2π,0).

Sub-Problem D
$\begin{align*} f(x,y)&=e^{-x} \cos{y}\\ \\ f_x&=-e^{-x}\cos{y}\\ 0&=-e^{-x}\cos{y}\\ y&=\frac{\pi}{2}+m\pi,\text{where }m \in \mathbb{Z} \\ \\ f_y&=-e^{-x} \sin{y}\\ y&=n\pi,\text{where }n \in \mathbb{Z} \\ \end{align*}$

We can see that the value of x such that f_x=0 and f_y=0 does not exist. So, this function does not have any critical points.

Sub-Problem E
$\begin{align*} f(x,y)&=\frac{1}{2}x^4 - 2x^3 + 4xy + y^2\\ \\ f_y&=2y+4x\\ -4x&=2y\\ -2x&=y\\ x&=-\frac{y}{2}\\ \\ f_x&=2x^3-6x^2+4y\\ 0&=x^3-3x^2+2y\\ &=x^3-3x^2-4x\\ &=x(x-4)(x+1)\\ \\ f_{xx} &= 6x^2-12x\\ f_{xy} &= 4\\ f_{yy} &= 2\\ \\ D(0,0) &= (6x^2-12x)(2) - (4)^2\\ &=-16\\ D(4,-8) &= (6x^2-12x)(2) - (4)^2\\ &=(96-48)(2) - 16\\ &=80\\ D(-1,2) &= (6+12)(2)-16\\ &=20\\ \end{align*}$

We have three critical points:

• Since D(0,0) < 0, it can be concluded that saddle point occurs during x=0 and y=0. The saddle point will be (0,0,0).
• Since D(4,-8) > 0 and f_xx(4,8) > 0, it can be concluded that relative minimum occurs during x=4 and y=8, where f(4,8)=-64
• Since D(-1,2) < 0 and f_xx(-1,2) > 0, it can be concluded that relative minimum also occurs during x=-1 and y=2, where f(-1,2)=-3,5.

Problem 7: Finding Tangent and Normal Lines
Given a function f and a point, find the tangent line and the normal line at the given point.

$\begin{align*} \text{A. } &2x + y - e^{x-y} = 2\;\;\;(1,1)\\ \text{B. } &z = \tan^{-1}\left ( \frac{y}{x} \right )\;\;\;(1,1,\frac{\pi}{4})\\ \text{C. } &\sin{xy}+3z=3 \;\;\;(0,3,1)\\ \text{D. } &z = xe^y \;\;\;(2,0,2)\\ \text{E. } &xyz = -4 \;\;\;(2,-1,2)\\ \end{align*}$

Sub-Problem A
$\begin{align*} &2x + y - e^{x-y} = 2\\ \\ &2x + y - e^{x-y} - 2 = 0\\ \\ f_x &= 2 - e^{x-y}\\ &=2 - e^0\\ &=1\\ f_y &= 1 + e^{x-y}\\ &=1+1\\ &=2\\ \\ f_x(x-a) + f_y(y-b)&=0\\ (x-1) + 2(y-1)&=0\\ x+2y-3&=0\\ \\ (x,y)&=(1,1)+ \lambda (1,2)\\ x&=1+\lambda\\ x-1&=\lambda\\ y&=1+2\lambda\\ \frac{y-1}{2}&=\lambda\\ x-1&=\frac{y-1}{2}\\ 2x-y-1&=0\\ \end{align*}$

Conclusion:

• Tangent line: x+2y-3=0
• Normal line: 2x-y-1=0

Sub-Problem B
$\begin{align*} &z = \tan^{-1}\left ( \frac{y}{x} \right )\\ \\ &0 = \tan^{-1}\left ( \frac{y}{x} \right ) - z\\ \\ f_x &= \frac{-\frac{y}{x^2}}{\frac{y^2}{x^2}+1}\\ &=-\frac{1}{2}\\ f_y &= \frac{\frac{1}{x}}{\frac{y^2}{x^2}+1}\\ &=\frac{1}{2}\\ f_z &= -1\\ \\ f_x(x-a) + f_y(y-b) + f_z(z-c)&=0\\ -\frac{1}{2}(x-1) + \frac{1}{2}(y-1) - \left ( z - \frac{\pi}{4} \right ) &= 0\\ -x+1+y-1-2z+\frac{\pi}{2}&=0\\ y-x-2z&=-\frac{\pi}{2}\\ \\ (x,y,z)&= \left ( 1,1,\frac{\pi}{4} \right ) + \lambda \left ( -\frac{1}{2},\frac{1}{2},-1 \right )\\ x&= 1 - \frac{\lambda}{2}\\ -2(x-1) &= \lambda\\ y&= 1 + \frac{\lambda}{2}\\ 2(y-1) &= \lambda\\ z&= \frac{\pi}{4} - \lambda\\ -\left ( z-\frac{\pi}{4} \right ) &=\lambda\\ \\ -2(x-1)&=2(y-1)=-\left ( z - \frac{\pi}{4} \right )\\ x-1 &= 1-y = \frac{z-\frac{\pi}{4}}{2} \end{align*}$

Conclusion:

• Tangent line: 2y-2x-4z=π
• Normal line: 8(x-1)=8(1-y)=4z-π

Sub-Problem C
$\begin{align*} &\sin{(xy)}+3z=3\\ \\ &\sin{(xy)}+3z-3=0\\ \\ f_x&=y \cos{(xy)}\\ &=3\\ f_y&=x \cos{(xy)}\\ &=0\\ f_z&=3\\ \\ f_x(x) + f_y(y-3) + f_z(z-1) &= 0\\ 3x+3(z-1)&=0\\ x+z&=1\\ \\ (x,y,z)&=(0,3,1) + \lambda (3,0,3)\\ x&=3\lambda\\ y&=3\\ z&=1+3\lambda\\ \\ x&=z-1,\;\;\;y=3\\ \end{align*}$

Conclusion:

• Tangent line: x+z=1;
• Normal line: x=z-1, y=3;

Sub-Problem D
$\begin{align*} &z = xe^y\\ \\ &z - xe^y = 0\\ \\ f_x&=-e^y\\ &=-1\\ f_y&=-xe^y\\ &=-2\\ f_z&=1\\ \\ f_x(x-2) + f_y(y) + f_z(z-2) &= 0\\ -(x-2) -2y + (z-2) &= 0\\ -x+2 -2y +z-2&=0\\ -x-2y+z&=0\\ \\ (x,y,z)&=(2,0,2) + \lambda (-1,-2,1)\\ x&=2 - \lambda\\ y&=-\lambda\\ z&=2+\lambda\\ \\ 2-x&=-y=z-2\\ \end{align*}$

Conclusion:

• Tangent line: -x-2y+z=0
• Normal line: 2-x=-y=z-2

Sub-Problem E
$\begin{align*} &xyz=-4\\ \\ &xyz+4=0\\ \\ f_x&=yz\\ &=-2\\ f_y&=xz\\ &=4\\ f_z&=xy\\ &=-2\\ \\ f_x(x-2) + f_y(y+1) + f_z(z-2) &= 0\\ -2(x-2)+4(y+1)-2(z-2)&=0\\ (x-2)-2(y+1)+(z-2)&=0\\ x-2y+z-6=0\\ \\ (x,y,z)&=(2,-1,2) + \lambda (-2,4,-2)\\ x&=2 - 2\lambda\\ y&=-1 + 4\lambda\\ z&=2 - 2\lambda\\ \\ \frac{2-x}{2}&=\frac{y+1}{4}=\frac{2-z}{2} \end{align*}$

Conclusion:

• Tangent line: x-2y+z-6=0
• Normal line: 2(2-x)=y+1=2(2-z)