Introduction To Partial Differentiation

Intermediate MathematicsIntroduction to PartialDifferentiationR Horan & M LavelleThe aim of this document is to provide a short, selfassessment programme for students who wish to acquirea basic understanding of partial differentiation.Copyright c 2004 rhoran@plymouth.ac.uk , mlavelle@plymouth.ac.ukLast Revision Date: May 25, 2005Version 1.0

Table of Contents1.2.3.4.Partial Differentiation (Introduction)The Rules of Partial DifferentiationHigher Order Partial DerivativesQuiz on Partial DerivativesSolutions to ExercisesSolutions to QuizzesThe full range of these packages and some instructions,should they be required, can be obtained from our webpage Mathematics Support Materials.

Section 1: Partial Differentiation (Introduction)31. Partial Differentiation (Introduction)In the package on introductory differentiation, rates of changeof functions were shown to be measured by the derivative. Manyapplications require functions with more than one variable: the idealgas law, for example, ispV kTwhere p is the pressure, V the volume, T the absolute temperature ofthe gas, and k is a constant. Rearranging this equation askTp Vshows that p is a function of T and V . If one of the variables, say T ,is kept fixed and V changes, then the derivative of p with respect toV measures the rate of change of pressure with respect to volume. Inthis case, it is called the partial derivative of p with respect to V andwritten as p. V

Section 1: Partial Differentiation (Introduction)kT, find the partial derivatives of p:V(a) with respect to T ,(b) with respect to V .Example 1 If p Solution(a) This part of the example proceeds as follows:kT,V pk , TVwhere V is treated as a constant for this calculation.(b) For this part, T is treated as a constant. Thus1p kT kT V 1 ,V pkT kT V 2 2 . VVp 4

Section 1: Partial Differentiation (Introduction)5The symbol is used whenever a function with more than one variableis being differentiated but the techniques of partial differentiation areexactly the same as for (ordinary) differentiation. z zExample 2 Findandfor the function z x2 y 3 . x ySolutionz x2 y 3For the first part y 3 is treated as za constant and the derivative of 2xy 3 , xx2 with respect to x is 2x. zFor the second part x2 is treatedand x2 3y 2 , yas a constant and the derivative2 2of y 3 with respect to y is 3y 2 . 3x y . z zandfor each of the following functions. x y(Click on the green letters for solutions.)1(a) z x2 y 4 ,(b) z (x4 x2 )y 3 , (c) z y 2 sin(x).Exercise 1. Find

Section 2: The Rules of Partial Differentiation62. The Rules of Partial DifferentiationSince partial differentiation is essentially the same as ordinary differentiation, the product, quotient and chain rules may be applied. zExample 3 Findfor each of the following functions. xx y(a) z xy cos(xy) , (b) z , (c) z (3x y)2 .x ySolution(a) Here z uv, where u xy and v cos(xy) so the product ruleapplies (see the package on the Product and Quotient Rules).u u xThus xyand yandv v x cos(xy) y sin(xy) . z u v v u y cos(xy) xy 2 sin(xy) . x x x

Section 2: The Rules of Partial Differentiation7(b) Here z u/v, where u x y and v x y so the quotient ruleapplies (see the package on the Product and Quotient Rules).u u xThus z x x yand 1andv v x x y 1. u v u x xv2(x y) (x y)2y .(x y)2(x y)2v (c) In this case z (3x y)2 so z u2 where u 3x y, and thechain rule applies (see the package on the Chain Rule).z u2 andu 3x y z u 2u and 3. u xThus z z u 2(3x y)3 6(3x y) . x u x

Section 2: The Rules of Partial Differentiation8 z zandfor each of the following functions. x y(Click on the green letters for solutions.)Exercise 2. Find(a) z (x2 3x) sin(y),(d) z sin(x) cos(xy),(b) z cos(x),y52(e) z e(x y 2 ),(c) z ln(xy),(f) z sin(x2 y).Quiz If z cos(xy), which of the following statements is true?(a) z z , x y(b) z1 z , xx y(c)1 z z ,y x y(d)1 z1 z .y xx y

Section 3: Higher Order Partial Derivatives93. Higher Order Partial DerivativesDerivatives of order two and higher were introduced in the package onMaxima and Minima. Finding higher order derivatives of functionsof more than one variable is similar to ordinary differentiation. 2z32Example 4 Findif z e(x y ) . x2Solution First differentiate z with respect to x, keeping y constant,then differentiate this function with respect to x, again keeping y32constant.z e(x y )32 z 3x2 e(x y ) using the chain rule x(x3 y 2 ) 2z (3x2 ) (x3 y2 ))2 (e e 3xusing the product rule x2 x x23232 z 6xe(x y ) 3x2 (3x2 e(x y ) )2 x32 (9x4 6x)e(x y )

Section 3: Higher Order Partial Derivatives10 2z 2zand, when there are two variables there x2 y 2is also the possibility of a mixed second order derivative. 2z32Example 5 Findif z e(x y ) . x y 2z zSolution The symbolis interpreted as; in words, x y x yfirst differentiate z with respect to y, keeping x constant, then differentiate this function with respect to x, keeping y constant. (It is thisdifferentiation, first with respect to x and then with respect to y, thatleads to the name of mixed derivative.)32 zFirst with x constant 2ye(x y ) (using the chain rule.) y 2z (x3 y2 ) Second with y constant 2ye x y xIn addition to both3 3x2 2ye(x3 6x2 ye(x y 2 ) y 2 ).

Section 3: Higher Order Partial Derivatives11The obvious question now to be answered is: what happens if theorder of differentiation is reversed? 2z z32Example 6 Find if z e(x y ) . y x y xSolution32 zFirst with y constant 3x2 e(x y ) (using the chain rule). x 2z 2 (x3 y2 ) Second with x constant 3x e y x y3 2y3x2 e(x y 2 ) 2z. x yAs a general rule, when calculating mixed derivatives the order ofdifferentiation may be reversed without affecting the final result.3 6x2 ye(x y 2 )

Section 3: Higher Order Partial Derivatives12Exercise 3. Confirm the statement on the previous page by finding 2z 2zbothandfor each of the following functions, whose first x y y xorder partial derivatives have already been found in exercise 2. (Clickon the green letters for solutions.)cos(x)(a) z (x2 3x) sin(y),(b) z ,(c) z ln(xy),y522(d) z sin(x) cos(xy),(e) z e(x y ) , (f) z sin(x2 y).Notation For first and second order partial derivatives there is a f fcompact notation. Thuscan be written as fx andas fy . x y 2f 2fis written fxx whileis written fxy .Similarly2 x x yQuiz If z e y sin(x), which of the following is zxx zyy ?(a) e y sin(x),(b) 0,(c) e y sin(x), (d) e y cos(x).

Section 4: Quiz on Partial Derivatives134. Quiz on Partial DerivativesChoose the correct option for each of the following.Begin Quiz1. If z x2 3xy y 3 then(a) 2x 3y 3y 2 ,(c) 2x 3x , zis x(b) 2x 3x 3y 2 ,(d) 2x 3y .2. If w 1/r, where r2 x2 y 2 z 2 , then xwx ywy zwz is(a) 1/r ,(b) 1/r ,rx3. If u then uxx isy(c) 1/r2 ,(d) 1/r2 .1111(a) p, (b) p, (c) p, (d) p.333334 y x4 yx8 y x8 yx3End Quiz Score:Correct

Solutions to Exercises14Solutions to ExercisesExercise 1(a) To calculate the partial derivative zof the function xz x2 y 4 , the factor y 4 is treated as a constant: z x2 y 4 x2 y 4 2x(2 1) y 4 2xy 4 . x x x zSimilarly, to find the partial derivative, the factor x2 is treated yas a constant: z x2 y 4 x2 y 4 x2 4y (4 1) 4x2 y 3 . y y yClick on the green square to return

Solutions to Exercises15 zfor the function z (x4 x2 )y 3 , the xfactor y 3 is treated as a constant: z (x4 x2 )y 3 x4 x2 y 3 (4x3 2x)y 3 . x x x zTo find the partial derivativethe factor (x4 x2 ) is treated as a yconstant: z 3 (x4 x2 )y 3 (x4 x2 ) y 3(x4 x2 )y 2 . y y yClick on the green square to return Exercise 1(b) To calculate

Solutions to Exercises16 z11Exercise 1(c) If z y 2 sin(x) then to calculatethe y 2 factor is xkept constant: 11 z 1y 2 sin(x) y 2 (sin(x)) y 2 cos(x) . x x x zSimilarly, to evaluate the partial derivativethe factor sin(x) is ytreated as a constant: 11 1 z 1y 2 sin(x) y 2 sin(x) y 2 sin(x) . y y y2Click on the green square to return

Solutions to Exercises17Exercise 2(a) The function z (x2 3x) sin(y) can be written asz uv , where u (x2 3x) and v sin(y) . The partial derivativesof u and v with respect to the variable x are u v 2x 3 , 0, x xwhile the partial derivatives with respect to y are u v 0, cos(y) . y yApplying the product rule z u v (2x 3) sin(y) .v u x x x v z u v u (x2 3x) cos(y) . y y yClick on the green square to return

Solutions to Exercises18Exercise 2(b)cos(x)can be written as z cos(x)y 5 .y5Treating the factor y 5 as a constant and differentiating with respectto x:The function z zsin(x) sin(x)y 5 5 . xyTreating cos(x) as a constant and differentiating with respect to y: vcos(x). cos(x)( 5y 6 ) 5 yy6Click on the green square to return

Solutions to Exercises19Exercise 2(c) The function z ln(xy) can be rewritten as (see thepackage on logarithms)z ln(xy) ln(x) ln(y) .Thus the partial derivative of z with respect to x is z 1 (ln(x) ln(y)) ln(x) . x x xxSimilarly the partial derivative of z with respect to y is z 1 (ln(x) ln(y)) ln(y) . y y yyClick on the green square to return

Solutions to Exercises20Exercise 2(d) To calculate the partial derivatives of the functionz sin(x) cos(xy) the product rule has to be applied z x z y sin(x) sin(x)cos(xy) , x x cos(xy)sin(x) sin(x)cos(xy) . y y cos(xy)Using the chain rule with u xy for the partial derivatives of cos(xy) cos(xy) x cos(xy) y cos(u) u cos(u) u u sin(u)y y sin(xy) , x u sin(u)x x sin(xy) . yThus the partial derivatives of z sin(x) cos(xy) are z cos(xy) cos(x) y sin(x) sin(xy) , xClick on the green square to return z x sin(x) sin(xy) . y

Solutions to Exercises212Exercise 2(e) To calculate the partial derivatives of z e(xchain rule has to be applied with u (x2 y 2 ): z x z y u(e ) u u(e ) u y 2 )the u u eu, x x u u eu. y yThe partial derivatives of u (x2 y 2 ) are u (x2 ) 2x , x x u (y 2 ) 2y . y y2Therefore the partial derivatives of the function z e(x22 z u eu 2x e(x y ) , x x22 zu u e 2y e(x y ) . x xClick on the green square to return y 2 )are

Solutions to Exercises22Exercise 2(f ) Applying the chain rule with u x2 y the partialderivatives of the function z sin(x2 y) can be written as z x z y (sin(u)) u (sin(u)) u u u, cos(u) x x u u cos(u). y yThe partial derivatives of u x2 y are u x2 2x , x x u y 1. y yThus the partial derivatives of the function z sin(x2 y) are z u 2x cos(x2 y) , cos(u) x x z u cos(u) cos(x2 y) . y yClick on the green square to return

Solutions to Exercises23Exercise 3(a)From exercise 2(a), the first order partial derivatives ofz (x2 3x) sin(y) are z z (2x 3) sin(y) , (x2 3x) cos(y) . x yThe mixed second order derivatives are 2z z (x2 3x) cos(y) (2x 3) cos(y) , x y x y x 2z z ((2x 3) sin(y)) (2x 3) cos(y) . y x y x yClick on the green square to return

Solutions to Exercises24Exercise 3(b)From exercise 2(b), the first order partial derivatives of z are zsin(x) zcos(x) , 5, xy5 yy6so the mixed second order derivatives are sin(x) 2z zcos(x) 5 6 , 5 x y x y xy6y 2 z z sin(x)sin(x) 5 5 6 . y x y x yyycos(x)y5Click on the green square to return

Solutions to Exercises25Exercise 3(c)From exercise 2(c), the first order partial derivatives of z ln(xy)are z 1 z 1 , . x x y yThe mixed second order derivatives are 1 2z z 0, x y x y x y 2z z 1 0. y x y x y xClick on the green square to return

Solutions to Exercises26Exercise 3(d) From exercise 2(d), the first order partial derivativesof z sin(x) cos(xy) are z z cos(x) cos(xy) y sin(x) sin(xy) , x sin(x) sin(xy) . x yThe mixed second order derivatives are 2z z ( x sin(x) sin(xy)) x y x y x sin(x) sin(xy) x cos(x) sin(xy) xy sin(x) cos(xy) , 2z z (cos(x) cos(xy) y sin(x) sin(xy)) y x y x y x cos(x) sin(xy) sin(x) sin(xy) xy sin(x) cos(xy) .N.B. In the solution above a product of three functions has beendifferentiated. This can be done by using two applications of theproduct rule.Click on the green square to return

Solutions to Exercises27Exercise 3(e) From exercise 2(e), the first order partial derivatives22of z e(x y ) are2222 z z 2xe(x y ) , 2ye(x y ) . x yThe mixed second order derivatives are thus 22 2z z (x2 y2 ) 2ye 4xye(x y ) , x y x y x 2222 2z z 2xe(x y ) 4yxe(x y ) . y x y x yClick on the green square to return