Partial Derivatives

1 ECON 331 Multivariable CalculusPartial DerivativesSingle variable calculus is really just a special case of multivariable calculus . For thefunctiony=f(x), we assumed thatywas the endogenous variable,xwas the exogenousvariable and everything else was a parameter. For example, given the equationsy=a+bxory=axnwe automatically treateda,b,andnas constants and took the derivative of y with respectto x (dy/dx). However, what if we decided to treatxas a constant and take the derivativewith respect to one of the other variables? Nothing precludes us from doing this. Considerthe equationy=axwheredydx=aNow suppose wefind the derivative ofywith respect toa,but TREATxas the we just reversed the roles played byaandxin our Variable Case:letz=f(x, y),whichmeans zis a function ofxandy.

2 Inthiscasezis the endoge-nous (dependent) variable and bothxandyare the exogenous (independent) variables. Tomeasure the the effect of a change in a single independent variable (x or y) on the dependentvariable (z) we use what is known as thePARTIAL DERIVATIVE. The Partial derivativeof z with respect to x measures the instantaneous change in the function as x changes whileHOLDING y constant. Similarly, we would hold x constant if we wanted to evaluate theeffect of a change in y on z. Formally: z xis the Partial derivative ofzwith respect tox,treatingyas a written asfx. z yis the Partial derivative ofzwith respect toy,treatingxas a written symbol ( bent over lower case D) is called the Partial symbol.

3 It is interpretedin exactly the same way asdydxfrom single variable calculus . The symbol simply serves toremind us that there are other variables in the equation, but for the purposes of the currentexercise, these other variables are held :z=x+y z/ x=1 z/ y=1z=xy z/ x=y z/ y=xz=x2y2 z/ x=2(y2)x z/ y=2(x2)yz=x2y3+2x+4y z/ x=2xy3+2 z/ y=3x2y2+4 REMEMBER:When you are taking a Partial derivative you treat the other variablesin the equation as constants! rules of Partial DifferentiationProduct Rule: givenz=g(x, y) h(x, y) z x=g(x, y) h x+h(x, y) g x z y=g(x, y) h y+h(x, y) g yQuotient Rule: givenz=g(x,y)h(x,y)andh(x, y)6=0 z x=h(x,y) g x g(x,y) h x[h(x,y)]2 z y=h(x,y) g y g(x,y) h y[h(x,y)]2 Chain Rule: givenz=[g(x, y)]n z x=n[g(x, y)]n 1 g x z y=n[g(x, y)]n 1 g yFurther Examples:For the functionU=U(x, y)find the the Partial derivates with respect to x and yfor each of the following examplesExample 1U= 5x3 12xy 6y5 Answer: U x=Ux=15x2 12y U y=Uy= 12x 30y42 Example 2U=7x2y3 Answer: U x=Ux=14xy3 U y=Uy=21x2y2 Example 3U=3x2(8x 7y)Answer.

4 U x=Ux=3x2(8) + (8x 7y)(6x)=72x2 42xy U y=Uy=3x2( 7) + (8x 7y)(0) = 21x2 Example 4U=(5x2+7y)(2x 4y3)Answer: U x=Ux=(5x2+7y)(2) + (2x 4y3)(10x) U y=Uy=(5x2+7y)( 12y2)+(2x 4y3)(7)Example 5U=9y3x yAnswer: U x=Ux=(x y)(0) 9y3(1)(x y)2= 9y3(x y)2 U y=Uy=(x y)(27y2) 9y3( 1)(x y)2=27xy2 18y3(x y)2 Example 6U=(x 3y)3 Answer: U x=Ux=3(x 3y)2(1) = 3(x 3y)2 U y=Uy=3(x 3y)2( 3) = 9(x 3y)23A Special Function: Cobb-DouglasThe Cobb-douglas function is a mathematical function that is very popular in economicmodels. The general form isz=xayband its Partial Derivatives are z/ x=axa 1yband z/ y=bxayb 1 Furthermore, the slope of the level curve of a Cobb-douglas is given by z/ x z/ y=MRS=abyxDifferentialsGiven the functiony=f(x)the derivative isdydx=f0(x)However, we can treatdy/dxas a fraction and factor out thedxdy=f0(x)dxwheredyanddxare be interpreted as the slope of afunction , thendyis the rise anddxis the run.

5 Another way of looking at it is asfollows: dy= the change iny dx= the change inx f0(x)= how the change inxcauses a change inyExample 7ify=x2thendy=2xdxLets supposex=2anddx= is the change iny(dy)?dy= 2(2)( ) = , atx=2,ifxis increased by thenywill increase by two variable caseIfz=f(x, y)then the change inzisdz= z xdx+ z ydy or dz=fxdx+fydywhichisreadas thechangeinz(dz)is due partially to a change inx(dx)plus partiallydue to a change iny(dy). For example, ifz=xythen the total differential isdz=ydx+xdyand, ifz=x2y3thendz=2xy3dx+3x2y2dyREMEMBER:Wh en you are taking the total differential, you are just taking all thepartial Derivatives and adding them 8 Find the total differential for the following utility (x1,x2)=ax1+bx2(a, b >0) (x1,x2)=x21+x32+ (x1,x2)=xa1xb2 Answers:1.

6 U x1=U1=a U x2=U2=bdU=U1dx1+U2dx2=adx1+bdx22. U x1=U1=2x1+x2 U x2=U2=3x22+x1dU=U1dx1+U2dx2=(2x1+x2)dx1+ (3x22+x1)dx23. U x1=U1=axa 11xb2=axa1xb2x1 U x2=U2=bxa1xb 12=bxa1xb2x2dU= axa1xb2x1 dx1+ bxa1xb2x2 dx2=hadx1x1+bdx2x2ixa1xb25 The Implicit Function TheoremSuppose you have a function of the formF(y, x1,x2)=0where the Partial Derivatives are F/ x1=Fx1, F/ x2=Fx2and F/ y= of functions are known as implicit functions whereF(y, x1,x2)=0implicity definey=y(x1,x2). What this means is that it is possible (theoretically) to rewrite to getyisolated and expressed as a function ofx1andx2. While it may not be possible to explicitlysolve for y as a function of x, we can stillfind the effect on y from a change inx1orx2byapplying the implicit function theorem:Theorem 9If a functionF(y, x1,x2)=0has well defined continuous Partial Derivatives F y=Fy F x1=Fx1 F x2=Fx2and if, at the values where F is being evaluated, the condition that F y=Fy6=0holds, thenyis implicitly defined as a function ofx.

7 The Partial Derivatives ofywith respecttox1andx2, are given by the ratio of the Partial Derivatives of F, or y xi= FxiFyi=1,2To apply the implicit function theorem tofind the Partial derivative ofywith respect tox1(for example),first take the total differential of FdF=Fydy+Fx1dx1+Fx2dx2=0then set all the differentials except the ones in question equal to zero ( setdx2=0)whichleavesFydy+Fx1dx1=0orFydy= Fx1dx1dividing both sides byFyanddx1yieldsdydx1= Fx1 Fywhich is equal to y x1from the implicit function 10 For eachf(x, y)=0,finddy/dxfor each of the 6x+7=0 Answer:dydx= fxfy= ( 6)1= +12x+17=0 Answer:dydx= fxfy= ( 12)3= +6x 13 y=0 Answer:dydx= fxfy= (2x+6) 1=2x+ (x, y)=3x2+2xy+4y3 Answer:dydx= fxfy= 6x+2y12y2+ (x, y)=12x5 2yAnswer:dydx= fxfy= 60x4 2= (x, y)=7x2+2xy2+9y4 Answer:dydx= fxfy= 14x+2y236y3+4xyExample 11 Forf(x, y, z)use the implicit function theorem tofinddy/dxanddy/ (x, y, z)=x2y3+z2+xyzAnswer:dydx= fxfy= 2xy3+yz3x2y2+xzdydz= fzfy= 2z+xy3x2y2+ (x, y, z)=x3z2+y3+4xyzAnswer:dydx= fxfy= 3x2z2+4yz3y2+4xzdydz= fzfy= 2x3z+4xy3y2+ (x, y, z)=3x2y3+xz2y2+y3zx4+y2zAnswer:dydx= fxfy= 6xy3+z2y2+4y3zx39x2y2+2xz2y+3y2zx4+2yzdy dz= fzfy= 2xzy2+y3x4+y29x2y2+2xz2y+3y2zx4+2yz8