2.3 Further exercises

Here are some further exercises to end this section.

Exercise 19

Calculate the partial derivatives prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis and prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis of each of the following functions.

  1. u of x comma y equals sum with 3 summands three times x plus x times y plus two times x squared times y squared

  2. u of x comma y equals x times cosine of y plus exp of x times y

  3. u of x comma y equals left parenthesis x plus y right parenthesis cubed

Answer

  1. Differentiating u of x comma y equals sum with 3 summands three times x plus x times y plus two times x squared times y squared with respect to x while keeping y fixed, we obtain

    prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals sum with 3 summands three plus y plus four times x times y squared full stop

    Differentiating with respect to y while keeping x fixed, we obtain

    prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals x plus four times x squared times y full stop
  2. Here u of x comma y equals x times cosine of y plus exp of x times y, so

    multiline equation row 1 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals cosine of y plus y times exp of x times y and row 2 prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals negative x times sine of y plus x times exp of x times y full stop
  3. Here u of x comma y equals left parenthesis x plus y right parenthesis cubed, so

    multiline equation row 1 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals three times left parenthesis x plus y right parenthesis squared and row 2 prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals three times left parenthesis x plus y right parenthesis squared full stop

Exercise 20

Calculate the partial derivatives prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis and prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis of each of the following functions, and evaluate these partial derivatives at left parenthesis one comma zero right parenthesis.

  1. u of x comma y equals x cubed times y minus y times cosine of y

  2. u of x comma y equals y times e super x minus x times y cubed

Answer

  1. Here u of x comma y equals x cubed times y minus y times cosine of y, so

    multiline equation row 1 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals three times x squared times y and row 2 prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals x cubed minus cosine of y plus y times sine of y full stop

    So, at left parenthesis x comma y right parenthesis equals left parenthesis one comma zero right parenthesis the partial derivatives have the values

    prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis one comma zero right parenthesis equals zero and prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis one comma zero right parenthesis equals zero full stop
  2. Here u of x comma y equals y times e super x minus x times y cubed, so

    multiline equation row 1 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals y times e super x minus y cubed and row 2 prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals e super x minus three times x times y squared full stop

    So, at left parenthesis x comma y right parenthesis equals left parenthesis one comma zero right parenthesis the partial derivatives have the values

    prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis one comma zero right parenthesis equals zero and prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis one comma zero right parenthesis equals e full stop

Exercise 21

Find the gradient of the graph of u of x comma y equals x squared plus two times x times y at the point left parenthesis one comma two comma five right parenthesis in the x-direction and in the y-direction.

Answer

Since u of x comma y equals x squared plus two times x times y, it follows that

prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals two times x plus two times y and prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals two times x full stop

The gradient of the graph at left parenthesis x comma y right parenthesis equals left parenthesis one comma two right parenthesis in the x-direction is

equation sequence part 1 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis one comma two right parenthesis equals part 2 two multiplication one plus two multiplication two equals part 3 six full stop

The gradient of the graph at left parenthesis x comma y right parenthesis equals left parenthesis one comma two right parenthesis in the y-direction is

equation sequence part 1 prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis one comma two right parenthesis equals part 2 two multiplication one equals part 3 two full stop

Exercise 22

Use the Cauchy–Riemann equations to show that there is no point of double-struck cap c at which the function

f times left parenthesis x plus i times y right parenthesis equals e super x times left parenthesis sine of y plus i times cosine of y right parenthesis

is differentiable.

Answer

Writing f in the form

f times left parenthesis x plus i times y right parenthesis equals u of x comma y plus i times v of x comma y comma

we obtain

u of x comma y equals e super x times sine of y and v of x comma y equals e super x times cosine of y full stop

Hence

multiline equation row 1 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals e super x times sine of y comma prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals e super x times cosine of y comma row 2 prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals e super x times cosine of y comma prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals negative e super x times sine of y full stop

If f is differentiable at x plus i times y, then the Cauchy–Riemann equations require that

multiline equation row 1 e super x times sine of y equals negative e super x times sine of y and row 2 e super x times cosine of y equals negative e super x times cosine of y semicolon

that is,

e super x times sine of y equals zero and e super x times cosine of y equals zero full stop

But e super x is never zero, so equation sequence part 1 sine of y equals part 2 cosine of y equals part 3 zero, which is impossible. It follows that there is no point of double-struck cap c at which f is differentiable.

Exercise 23

Use the Cauchy–Riemann equations to show that the function

f times left parenthesis x plus i times y right parenthesis equals left parenthesis x squared plus x minus y squared right parenthesis plus i times left parenthesis two times x times y plus y right parenthesis

is entire, and find its derivative.

Answer

In this case,

multiline equation row 1 u of x comma y equals x squared plus x minus y squared and row 2 v of x comma y equals two times x times y plus y comma

so

multiline equation row 1 Blank prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals two times x plus one comma prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals two times y comma row 2 Blank prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals negative two times y comma prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals two times x plus one full stop

These partial derivatives are defined and continuous on the whole of double-struck cap c. Furthermore,

multiline equation row 1 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis and row 2 prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals negative prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis comma

so the Cauchy–Riemann equations are satisfied at every point of double-struck cap c.

By the Cauchy–Riemann Converse Theorem, f is entire, and

multiline equation row 1 f super prime times left parenthesis x plus i times y right parenthesis equals prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis plus i times prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis row 2 Blank equals left parenthesis two times x plus one right parenthesis plus two times y times i full stop

(So f super prime of z equals two times z plus one, and in fact f of z equals z squared plus z.)

Exercise 24

Use the Cauchy–Riemann equations to find all the points at which the following functions are differentiable, and calculate their derivatives.

  1. f times left parenthesis x plus i times y right parenthesis equals left parenthesis x squared plus y squared right parenthesis plus i times left parenthesis x squared minus y squared right parenthesis

  2. f times left parenthesis x plus i times y right parenthesis equals x times y

Answer

  1. Here

    u of x comma y equals x squared plus y squared and v of x comma y equals x squared minus y squared comma

    so

    multiline equation row 1 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals two times x comma prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals two times x comma row 2 prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals two times y comma prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals negative two times y full stop

    The Cauchy–Riemann equations are satisfied only if x equals negative y. So f cannot be differentiable at x plus i times y unless x equals negative y. Since the partial derivatives above exist, and are continuous on double-struck cap c (and in particular when x equals negative y), it follows from the Cauchy–Riemann Converse Theorem that f is differentiable on the set x plus i times y colon x equals negative y.

    On this set,

    multiline equation row 1 f super prime times left parenthesis x plus i times y right parenthesis equals prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis plus i times prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis row 2 Blank equation sequence part 1 equals part 2 two times x plus two times x times i equals part 3 two times x times left parenthesis one plus i right parenthesis full stop
  2. Here

    u of x comma y equals x times y and v of x comma y equals zero comma

    so

    multiline equation row 1 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals y comma prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals zero comma row 2 prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals x comma prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals zero full stop

    The Cauchy–Riemann equations are not satisfied unless y equals zero and negative x equals zero. So f is not differentiable except possibly at 0. Since the partial derivatives above exist, and are continuous at open zero comma zero close, it follows from the Cauchy–Riemann Converse Theorem that f is differentiable at 0. Furthermore,

    equation sequence part 1 f super prime of zero equals part 2 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis zero comma zero right parenthesis plus i times prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis zero comma zero right parenthesis equals part 3 zero full stop