3.2 Higher Order Partial Derivatives

1 Higher Order Partial DerivativesIffis a function of several variables, then we can find Higher Order partialsin the following (x,y)is a function of two variables, then f xand f yarealso functions of two variables and their partials can be taken. Hence we candifferentiate them with respect toxandyagain and find, 2f x2, the derivative offtaken twice with respect tox, 2f x y, the derivative offwith respect toyand then with respect tox, 2f y x, the derivative offwith respect toxand then with respect toy, 2f y2, the derivative offtaken twice with respect can carry on and find 3f x y2, which is taking the derivative offfirst withrespect toytwice, and then differentiating with respect tox, etc.

2 In thismanner we can findnth- Order Partial Derivatives of a 2f x yand 2f y xare called mixed Partial Derivatives . They are equalwhen 2f x yand 2f y xare this course all the fuunctions we will encounter will have equalmixed Partial Find all partials up to the second Order of the functionf(x,y) =x4y2 : f x= 4x3y2 2xy6 f y= 2x4y 6x2y5 2f x2= 12x2y2 2y6 2f y x= 8x3y 12xy5 2f y2= 2x4 30x2y4 2f x y= 8x3 12xy5 Notations:fx= f xfy= f yfxx= 2f x2fyy= 2f y2fxy= 2f x Chain RuleYou are familiar with the chain rule for functions of one variable: iffisa function ofu, denoted byf=f(u), anduis a function ofx, denotedu=u(x).

3 Thendfdx= Rules for First- Order Partial DerivativesFor a two-dimensional version, supposezis a function ofuandv, denotedz=z(u,v)anduandvare functions ofxandy,u=u(x,y) andv=v(x,y)then z x= z u u x+ z v v y z y= z u u y+ z v v yz z uwwoooooooooooooo z v''PPPPPPPPPPPPPPu u x u y ????????v v x v y ??????? Find the first Partial Derivatives using chain (u,v)u=xyv= 2x+ 3ySolution: z x= zu u x+ zv v x=y zu+ 2 zv z y= zu u y+ zv v y=x zu+ 3 zvChain Rule for Second Order Partial DerivativesTo find second Order partials, we can use the same techniques as first orderpartials, but with more care and patience!

4 (u,v)u=x2yv= 3x+ 2y1. Find 2z :We will first find 2z y2. z y= z u u y+ z v v y=x2 z u+ 2 z differentiate again with respect toyto obtain 2z y2= y(x2 z u) + y(2 z v)=x2 y( z u) + 2 y( z v)Note thatzis a function ofuandv, anduandvare functions ofxandy. Then the Partial Derivatives z uand z vare also initially functions ofuandvand eventually functions ofxandy. In other words we havethe same dependence diagram y( z u)6= 2z y u). Never write 2z y uas it is mathematically mean-ingless. z u 2z u2wwpppppppppppppp 2z v u''NNNNNNNNNNNNNNu u x u y ?

5 ???????v v x v y ???????xyxy z v 2z u vwwpppppppppppppp 2z v2''NNNNNNNNNNNNNNu u x u y ????????v v x v y ???????xyxyTherefore 2z y2=x2( 2z u2 u y+ 2z v u v y) + 2( 2z u v u y+ 2z v2 v y)=x2(x2 2z u2+ 2 2z v u) + 2(x2 2z u v+ 2 2z v2)The mixed partials are equal so the answer simplifies to 2z y2=x4 2z u2+ 4x2 2z u v+ 4 2z Find 2z x ySolution:We have z y= z u u y+ z v v y=x2 z u+ 2 z differentiate again with respect toxto obtain 2z x y= x(x2 z u) + x(2 z v)=x2 x( z u) + 2x z u+ 2 x( z v)=x2( 2z u2 u x+ 2z v u v x) + 2x z u+ 2( 2z u v u x+ 2z v2 v x)=x2(2xy 2z u2+ 3 2z v u) + 2x z u+ 2(2xy 2z u v+ 3 2z v2)= 2x3y 2z u2+ (3x2+ 4xy) 2z v u) + 2x z u+ 6 2z Maxima and MinimaRecall from 1-dimensional calculus, to find the points of maxima and minimaof a function, we first find the critical points where the tangent line ishorizontalf (x) = 0.

6 Then(i) Iff (x)>0 the gradient is increasing and we have a local minimum.(ii) Iff (x)<0 the gradient is decreasing and we have a local maximum.(iii) Iff (x) = 0 it is Points of a function of 2 variablesWe are now interested in finding points of local maxima and minima for afunction of two functionf(x,y)has a relative minimum (resp. maximum)at the point(a,b)iff(x,y) f(a,b)( (x,y) f(a,b)) for all points(x,y)in some region around(a,b) point(a,b)is a critical point of a functionf(x,y)if one ofthe following is true(i)fx(a,b) = 0andfy(a,b) = 0(ii)fx(a,b)and/orfy(a,b)does not of Critical PointsWe will need two quantities to classify the critical points off(x,y) , the second Partial derivative offwith respect f2xythe HessianIf the Hessian is zero, then the critical point is degenerate.

7 If the Hessianis non-zero, then the critical point is non-degenerate and we can classify thepoints in the following manner:case(i) IfH>0 andfxx<0 then the critical point is a relative (ii) IfH>0 andfxx>0 then the critical point is a relative (iii) IfH<0 then the critical point is a saddle and classify the critical points of12x3+y3+ 12x2y :We first find the critical points of the 36x2+ 24xy= 12x(3x+ 2y)fy= 3y2+ 12x2 75 = 3(4x2+y2 25).The critical points are the points wherefx= 0 andfy= 0fx= 012x(3x+ 2y) = 0 Therefore eitherx= 0 or 3x+ 2y= 0.

8 We handle the two cases separately;case(i)x= substituting this infywe getfy= 3(y2 25) = 0 impliesy= (ii) 3x+ 2y= 3x/2 and substituting this infywe get,fy= 3(4x2+9x24 25)=34(16x2+ 9x2 100)=34(25x2 100)=754(x2 4)fy= 0754(x2 4) = 0x2 4 = 0x= 2 Thus we have found four critical points: (0,5),(0, 5),(2, 3),( 2,3).We must now classify these 72x+ 24y= 24(3x+y)fxy= 24xfyy= 6yH=fxxfyy f2xy= (24)(3x+y)(6y) (24x)2 PointsfxxHType(0,5)1203600 Minimum(0, 5)-1203600 Maximum(2, 3)72-3600 Saddle( 2,3)-72-3600 SaddleReferences1.

9 Engineering Mathematics, by K. A. A complete set of notes on Pre-Calculus, Single Variable Calculus, Multi-variable Calculus and Linear is a link to the chapter on Higher Order Partial is a link to the chapter on Chain is a link to the chapter on Maxima and A collection of examples, animations and notes on Multivariable jflores/MultiCalc02/WebBook/Chapter15/4. Links to various resources on