Answer :
The value of ∂5f/∂x2∂y3 at (1,1) is 15e+16e+12e+564= 593.53
Given function is f(x,y)=xey2/2+94x2y3.
To find ∂5f/∂x2∂y3 at (1,1)Let's first find the higher order partial derivative ∂5f/∂x2∂y3.
Therefore, we differentiate the function four times with respect to x and three times with respect to y
∂5f/∂x2∂y3=fifth partial derivative of f(x,y)=xey2/2+94x2y3∂/∂x[f(x,y)]
= ∂/∂x[xey2/2+94x2y3]
= y2e^(y2/2)+ 846y3x∂2f/∂x2
= ∂/∂x[y2e^(y2/2)+ 846y3x]
= 94y3+ 6768y3x∂3f/∂x3
= ∂/∂x[94y3+ 6768y3x]= 0∂4f/∂x4= ∂/∂x[0]
= 0∂/∂y[f(x,y)]= ∂/∂y[xey2/2+94x2y3]
= xy*e^(y2/2)+ 282x2y2∂2f/∂y2
= ∂/∂y[xy*e^(y2/2)+ 282x2y2]
= x(e^(y2/2)+ 2y2e^(y2/2))+ 564xy∂3f/∂y3
= ∂/∂y[x(e^(y2/2)+ 2y2e^(y2/2))+ 564xy]
= x(3y*e^(y2/2)+ 2ye^(y2/2)+ 4y3e^(y2/2))+ 564x∂4f/∂y4
= ∂/∂y[x(3y*e^(y2/2)+ 2ye^(y2/2)+ 4y3e^(y2/2))+ 564x]
= x(15y2e^(y2/2)+ 16ye^(y2/2)+ 12y4e^(y2/2))+ 564∂5f/∂x2∂y3
= ∂/∂x[x(15y2e^(y2/2)+ 16ye^(y2/2)+ 12y4e^(y2/2))+ 564]
= 15y2e^(y2/2)+ 16ye^(y2/2)+ 12y4e^(y2/2)+ 564
∴ The value of ∂5f/∂x2∂y3 at (1,1) is 15e+16e+12e+564= 593.53 (approx).Hence, the answer is 593.53.
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