経済学で出る数学

ワークブックでじっくり攻める：問7.2解答例

1. • $\dfrac{\partial f}{\partial x}(x,y)=3\times 2xy^4=6xy^4$
   
   
   • $\dfrac{\partial f}{\partial y}(x,y)=3x^2\times(4y^3)=12x^2y^3$
   
   


2. • $\dfrac{\partial f}{\partial x}(x,y)=4\times 3x^2y=12x^2y$
   
   
   • $\dfrac{\partial f}{\partial y}(x,y)=4x^3\times 1=4x^3$
   
   


3. • $\dfrac{\partial f}{\partial x}(x,y)=1\times y +1+0=y+1$
   
   
   • $\dfrac{\partial f}{\partial y}(x,y)=x\times 1 +0+1=x+1$
   
   


4. • $\dfrac{\partial f}{\partial x}(x,y)=4x+2\times 1\times y+0-6-0 =4x+2y-6$
   
   
   • $\dfrac{\partial f}{\partial y}(x,y)=0+2x\times 1+2y-0-4=2x+2y-4$
   
   


5. • $\dfrac{\partial f}{\partial x}(x,y)=2x+4\times 1 \times y+0-2+0+0 =2x+4y-2$
   
   
   • $\dfrac{\partial f}{\partial y}(x,y)=0+4x\times 1+18y-0+6-0=4x+18y+6$
   
   


6. • $\dfrac{\partial f}{\partial x}(x,y)=3x^2-0-3+0=3x^2-3$
   
   
   • $\dfrac{\partial f}{\partial y}(x,y)=0-3y^2-0+12=-3y^2+12$
   
   


7. • $\dfrac{\partial f}{\partial x}(x,y)=\left(\alpha x^{\alpha-1}\right)y^{\beta}=\alpha x^{\alpha-1}y^{\beta}$
   
   
   • $\dfrac{\partial f}{\partial y}(x,y)=x^{\alpha}\left(\beta y^{\beta-1}\right)= \beta x^{\alpha}y^{\beta-1}$
   
   


8. • $\dfrac{\partial f}{\partial x}(x,y) =\left(\dfrac{1}{2} x^{-\frac{1}{2}}\right)y^{\frac{1}{2}} =\dfrac{1}{2} x^{-\frac{1}{2}}y^{\frac{1}{2}}$
   
   
   • $\dfrac{\partial f}{\partial y}(x,y)= x^{\frac{1}{2}}\left(\dfrac{1}{2} y^{-\frac{1}{2}}\right) =\dfrac{1}{2} x^{\frac{1}{2}}y^{-\frac{1}{2}}$
   
   


9. • $\dfrac{\partial f}{\partial x}(x,y) =\left(\dfrac{1}{3} x^{-\frac{2}{3}}\right)y^{\frac{2}{3}} =\dfrac{1}{3} x^{-\frac{2}{3}}y^{\frac{2}{3}}$
   
   
   • $\dfrac{\partial f}{\partial y}(x,y)= x^{\frac{1}{3}}\left(\dfrac{2}{3} y^{-\frac{1}{3}}\right) =\dfrac{2}{3} x^{\frac{1}{3}}y^{-\frac{1}{3}}$
   
   


10. • $\dfrac{\partial f}{\partial x}(x,y)=e^x+1\times \log_{}{y}= e^x+\log_{}{y}$
    
    
    • $\dfrac{\partial f}{\partial y}(x,y)=0+x\times\dfrac{1}{y} =\dfrac{x}{y}$
    
    


11. • $\dfrac{\partial f}{\partial x}(x,y)=\dfrac{\alpha}{x}+0 =\dfrac{\alpha}{x}$
    
    
    • $\dfrac{\partial f}{\partial y}(x,y)=0+\dfrac{1-\alpha}{y} =\dfrac{1-\alpha}{y}$
    
    


12. • $\dfrac{\partial f}{\partial x}(x,y)=e^y\times \dfrac{1}{x} =\dfrac{e^y}{x}$
    
    
    • $\dfrac{\partial f}{\partial y}(x,y)=e^y\times \log_{}{x} =e^y\log_{}{x}$
    
    


13. • $\dfrac{\partial f}{\partial x}(x,y)= \dfrac{(xy)^{\prime}(x+y)-(xy)(x+y)^{\prime}}{(x+y)^2} =\dfrac{(y)(x+y)-(xy)(1+0)}{(x+y)^2} =\dfrac{yx+y^2-xy}{(x+y)^2}=\dfrac{y^2}{(x+y)^2}$
    
    
    • $\dfrac{\partial f}{\partial y}(x,y)= \dfrac{(xy)^{\prime}(x+y)-(xy)(x+y)^{\prime}}{(x+y)^2} =\dfrac{(x)(x+y)-(xy)(0+1)}{(x+y)^2} =\dfrac{x^2+xy-xy}{(x+y)^2}=\dfrac{x^2}{(x+y)^2}$$
    
    


14. • $\dfrac{\partial f}{\partial x}(x,y)= \dfrac{(2xy)^{\prime}(x+y)-(2xy)(x+y)^{\prime}}{(x+y)^2} =\dfrac{(2y)(x+y)-(2xy)(1+0)}{(x+y)^2} =\dfrac{2yx+2y^2-2xy}{(x+y)^2}=\dfrac{2y^2}{(x+y)^2}$
    
    
    • $\dfrac{\partial f}{\partial y}(x,y)= \dfrac{(2xy)^{\prime}(x+y)-(2xy)(x+y)^{\prime}}{(x+y)^2} =\dfrac{(2x)(x+y)-(2xy)(0+1)}{(x+y)^2} =\dfrac{2x^2+2xy-2xy}{(x+y)^2}=\dfrac{2x^2}{(x+y)^2}$$