Vector Calculus - Packet 09

Partial Derivatives

Suppose we are given a function $f:{ℝ}^2\to ℝ$, written in terms of components by $f(x,y)\in ℝ$ for each $(x,y)\in {ℝ}^2$. Now let the second coordinate remain some arbitrary fixed value $y=b$. This determines a function $f(x,b)$ of one variable, $x$, for each possible chosen $b$.

We can take the derivative of $f(x,y_0)$ in $x$ just as any other function of $x$:

$$\frac{d}{dx}f(x,b)=\underset{h\to 0}{\lim }\frac{f(x+h,b)-f(x,b)}{h}.$$

The resulting function of $x$ may be considered for various possible values of $b$, i.e. the variation may be restored. This function of both $x$ and $y$ is called the partial derivative of $f$ with respect to $x$. As with ordinary derivatives, a variety of notations is used:

$$f_x(x,y)=\frac{\partial }{\partial x}f(x,y)=\frac{\partial f}{\partial x}(x,y)=\underset{h\to 0}{\lim }\frac{f(x+h,y)-f(x,y)}{h}.$$

The new symbol ‘ $\partial$ ’ is called “partial” and should be written differently from ‘ $d$ ’. This symbol indicates the function’s dependence on additional variables.

Functions of several variables may have partial derivatives in each of their variables. Thus, for $f$ above we also have $f_y$ or $\frac{\partial f}{\partial y}$.

By interpreting the function $f(x,b)$ as the “trace” of the function $f(x,y)$ at a fixed value $y=b$, one sees that the partial derivative gives the slope of the curve formed as the cross-section of the plane $y=b$ with the graph of $f$, namely the surface $z=f(x,y)$: Similarly for the partial $f_y$ and the trace $x=a$:

One can iterate partial derivatives and consider higher partials in the obvious manner. The notation $f_{xy}$ indicates the “mixed partial” taken in $x$ first:

$$f_{xy}=\frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x}\right)=\frac{{\partial }^{2}f}{\partial y\partial x}.$$

In principle, the double partials $f_{xy}$ and $f_{yx}$ could be different, but for nice functions they do agree by Clairaut’s Theorem:

Clairaut’s Theorem

Provided $f_{xy}$ and $f_{yx}$ exist and are continuous on a region $ℛ$, then $f_{xy}=f_{yx}$ on that region.

(Caveat: in the theorem, $ℛ$ should be ‘open’ topologically: it should have the same dimension as the background and not include its boundary.)

Example

Partial derivatives of a monomial

Problem: Compute the partial derivatives of first and second order of the function $f(x,y)=x^3{y}^5$. Solution: We have $f_x=3x^2{y}^5$, $\frac{\partial f}{\partial y}=5x^3{y}^4$, $f_{xy}=f_{yx}=15x^2{y}^4$, $\frac{{\partial }^{2}f}{\partial x^2}=6x{y}^5$, $f_{yy}=20x^3{y}^3$.

Exercise 09A-01

Partial derivatives of a composite

Let $f(x,y)=\sin (\frac{x}{1+y})$. Compute $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$.

Exercise 09A-02

Evaluate partial with four variables

Let

$$f(x,y,z,w)=\frac{e^{xz+y}}{z^2+w}.$$

Calculate $f_z(\mathrm{0,0,1,1})$.

Exercise 09A-03

Higher order partials

Let $f(x,y,z)=x^3+y^2{e}^x$.

• (a) Calculate the second-order partial derivatives of $f$.
• (b) Calculate $f_{xyy}$.

Exercise 09A-04

Clairaut: choose the order wisely!

Let $f(x,y,z,w)=x^3{w}^2{z}^2+\sin (\frac{xy}{z^2})$. Compute $f_{zzwx}$ in your head. (Explain it on paper!)

Continuity and differentiability: more complex in 3d!

The concepts of continuity and differentiability are more complicated in 3d due to potential incompatibilities of directionality that are a function of the $x$ and $y$ relationship. $z=f(x,y)=\frac{xy}{x^2+y^2}$ $z=f(x,y)=\frac{2xy(x+y)}{x^2+y^2}$

Linear approximation

There is a relation between the concepts of linear approximation, partial derivative, tangent plane, and differentiability. This relation is contained in a function defined in terms of $f(x,y)$:

Linear approximation function

$$L(x,y)=f(a,b)+f_x(a,b)\cdot (x-a)+f_y(a,b)\cdot (y-b).$$

The linear approximation function includes both partial derivatives $f_x$ and $f_y$. The set of points satisfying $z=L(x,y)$ defined by the above equation, namely the graph of $L(x,y)$, is the tangent plane. Lastly, the function $f(x,y)$ is said to be differentiable when:

$$f(x,y)=L(x,y)+(x-a)\cdot {\epsilon }_1+(y-b)\cdot {\epsilon }_2,$$

with two functions ${\epsilon }_1$ and ${\epsilon }_2$ satisfying ${\epsilon }_1,{\epsilon }_2\to 0$ as $(x,y)\to (a,b)$.

If $f(x,y)$ is not differentiable, then the linear approximation may be a bad approximation.

Linear approximation in 2d

Remember that for a function $f(x)$ of a single variable, we have the linear approximation:

$$L(x)=f(a)+f^{\prime }(a)\cdot (x-a).$$

The tangent line to $f$ at $x=a$ is given by $y=L(x)$, and $f$ is considered differentiable when $f(x)=L(x)+(x-a)\cdot \epsilon$ such that $\epsilon \to 0$ as $x\to a$:

$$(\frac{f(x)-f(a)}{x-a}-f^{\prime }(a))=\epsilon \to 0\text{ as }x\to a\text{iff }{f}^{\prime }(x)\text{ exists at }a.$$

Tangent plane via tangent lines

How do we know that the graph of $L(x,y)$ gives the tangent plane?

The tangent vector to the trace $z=f(x,b)$ at $x=a$ is given by ${\frac{d}{dx}(x,b,f(t,b))|}_a=(\mathrm{1,0},f_x(a,b))$. So the tangent line is given by:

$${𝐪}_{(a,b)}(t)=(a,b,f(a,b))+t(\mathrm{1,0},f_x(a,b))=(a+t,b,f(a,b)+t{f}_x(a,b)).$$

These points satisfy $z=L(x,y)$. The graph $z=L(x,y)$ must be the tangent plane because it is the unique plane such that the tangent lines of the trace curves $f(x,b)$ and $f(a,y)$ lie in this plane.

Taylor series and linear approximation

Linear approximation gives the first two terms of the Taylor series of a function, and the remainder term, captured in $\epsilon$, must limit to zero when the function is differentiable:

$$\begin{array}{cc}f(x)& =f(a)+f^{\prime }(a)(x-a)+\frac{1}{2!}{f}^{\prime \prime }(a){(x-a)}^2+\dots \\ & =f(a)+f^{\prime }(a)(x-a)+\epsilon \cdot (x-a).\end{array}$$

If $f$ agrees with its Taylor series, then:

$$\epsilon (x)=\frac{1}{2!}{f}^{\prime \prime }(a)(x-a)+\frac{1}{3!}{f}^{(3)}(a){(x-a)}^2+\frac{1}{4!}{f}^{(4)}{(x-a)}^3+\dots$$

Differentials

Some authors use the mnemonic notation of finite differentials for applications of linear approximation. These are written as $dx$, $dy$, $dz$, $df$, etc., even though they are not infinitesimals. Their usage is according to the following kind of rules:

$$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy.$$

Example

Tangent plane to paraboloid

Problem: Find the equation of the plane tangent to the paraboloid $z=2x^2+y^2$ at $(x,y,z)=(\mathrm{1,1,3})$. Solution: Set $f(x,y)=2x^2+y^2$. Compute $f_x=4x$, $f_y=2y$. So $f_x(\mathrm{1,1})=4$, $f_y(\mathrm{1,1})=2$, and the linear approximation is given by:

$$L(x,y)=3+4(x-1)+2(y-1).$$

The tangent plane is found by setting $z=L(x,y)$.

Exercise 09B-01

Tangent plane to a graph

Let $f(x,y)=x{y}^3+x^2$. Find the tangent plane to the graph of $f(x,y)$ at the point $(x,y)=(2,-2)$.

Exercise 09B-02

Linear approximation

Use a linear approximation to estimate the value of ${(3.99)}^3{(1.01)}^4{(1.98)}^{-1}$.

• (a) Write this using a function $L(x,y,z)$.
• (b) Write this using the notation of differentials.

What is the percentage error of your estimate from the true value?

Differentiability

In 2d, a function $f(x)$ is differentiable precisely where its derivative $f^{\prime }(x)$ exists. The situation is not so simple in 3d: a function like $f(x,y)=\frac{2xy(x+y)}{x^2+y^2}$ has both partial derivatives at $(x,y)=(\mathrm{0,0})$, but it is not differentiable there! On the other hand, one theorem does help:

Theorem

If a function $f(x,y)$ is given and both partials $f_x$ and $f_y$ exist and are continuous at a point $(a,b)$, then $f$ is differentiable at $(a,b)$.

Exercise 09B-03

When partials aren’t continuous

Compute the partial derivatives of $f(x,y)=\frac{2xy(x+y)}{x^2+y^2}$ and explain carefully why this non-differentiable function does not violate the theorem.

Exercise 09B-04

Differentiability of the radius function

Let $r(x,y)=\sqrt{x^2+y^2}$. Is $r(x,y)$ differentiable? Where?

Differentiable, but partials aren’t continuous: $r^2\sin (\frac{1}{r})$

The function $f(r,\theta )=r^2\sin (\frac{1}{r})$ is differentiable at $(\mathrm{0,0})$ since it is approximated by its tangent plane at $(\mathrm{0,0})$. The partials are not continuous there! You may check this by computing $f_x$ and $f_y$. You can also compute $f_r=2r\sin (1/r)-\cos (1/r)$, which is not continuous at $r=0$.

Problems due 22 Oct 2023, 9:00pm

Problem 09-01

Partial derivatives

• (a) Find $\frac{\partial }{\partial y}(x^2+y)(x+y^4)$.
• (b) Find $\frac{\partial }{\partial u}\ln (u^2+uv)$.
• (c) Compute $f_{xyxzy}$ for $f(x,y,z)=y\sin (xz)\sin (x+z)+(x+z^2)\tan (y)+x\tan \left(\frac{z+z^{-1}}{y-y^{-1}}\right)$. (Hint: you can use a different order on each term!)
• (d) The Laplace Operator $\mathrm{\Delta }$ is defined by $\mathrm{\Delta }f=f_{xx}+f_{yy}$. It is very important! Show that $u(x,y)=e^x\cos (y)$ and $v(x,y)=\ln (x^2+y^2)$ and $\theta (x,y)={\tan }^{-1}(\frac{y}{x})$ are all annihilated (sent to 0) by $\mathrm{\Delta }$.

Problem 09-02

Linear approximation

Use linear approximation to estimate the value of $\frac{8.01}{\sqrt{(1.99)(2.01)}}$ without a calculator, using the notation of differentials. Then compute the exact value on a calculator and find the percentage error.

Problem 09-03

Tangent planes

• (a) Find the tangent plane to the graph of $f(x,y)=x{y}^3+x^2$ at the point $(2,-2,-12)$.
• (b) Suppose $f(\mathrm{1,2})=10$, $f(\mathrm{1.1,2.01})=10.3$ and $f(\mathrm{1.04,2.1})=9.7$. Find an approximate equation of the tangent plane to the graph $z=f(x,y)$ at the point $(\mathrm{1,2,10})$.
• (c) Find out where the tangent plane to $z=3x^2-4y^2$ has normal vector $(\mathrm{3,2,2})$. (Hint: by inspection of the equation $z=L(x,y)$, you can write a general formula for the normal vector to a tangent plane. Insert an arbitrary point $(a,b)$ and solve for $a$ and $b$ to be where this normal vector aligns with the given one.)

$★$ Problem 09-04

Please write up and submit exercises 09A-03, 09A-04, 09B-01, 09B-02, 09B-04.