- Second-Order Partial Derivatives
By differentiating a function z = f(x, y) twice , we get its second-order derivatives.
These derivatives are usually denoted by:
 |
1. If f( x, y) = x cos y + y e^x , then find
∂f/∂x , ∂f/∂y , ∂^2f /∂x^2 , ∂^2f/∂y^2 , ∂^2f/∂x∂y , ∂^2f/∂y∂x.
Solution:
∂f/∂x = ∂ / ∂x (x cos y + y e^x) = cos y + y e^x
∂f/∂y = ∂/∂y (x cos y + y e^x) = −x sin y + e^x
∂^2f/∂x^2 = ∂/∂x(∂f/∂x) = ∂ /∂x (cos y + y e^x) = 0 + y e^x = y e^x
∂^2f/∂y^2 = ∂/∂y (∂f/∂y) = ∂/∂y( −x sin y + e^x )= −x cos y + 0 = −x cos y
∂^2f/∂x∂y = ∂/∂x(∂f/∂y) = ∂/∂x (−x sin y + e^x) = − sin y + e^x
∂^2f/∂y∂x = ∂/∂y(∂f/∂x) = ∂/∂y (cos y + y e^x) = − sin y + e^x
2.If u = log x^2 + y^2 verify ∂^2u/∂x∂y = ∂^2u/ ∂y∂x
Solution:
∂u/∂x= 2x/x^2 + y^2
∂u/∂y = 2y/x^2 + y^2
∂^2u/∂x∂y = ∂/∂x(∂u/∂y) = ∂/∂x(2y/x^2 + y^2)
=(x^2+y^2) ∂/∂x (2y) − (2y) ∂/∂x x^2 + y^2
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(x^2 + y^2)^2
=(x^2+y^2)(0) − 2y(2x)
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(x^2 + y^2)^
= −4xy/(x^2 + y^2)^2
∂^2u/∂y∂x = ∂/∂y(∂u/∂x) = ∂/∂y(2x/x^2 + y^)2
= (x^2+y^2) ∂/∂y(2x) − 2x∂/∂y(x^2 + y^2)
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(x^2 + y^2)^2
=(x^2+y^2)(0) − 2x 2y
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(x^2 + y^2)^2
= − 4xy (x^2 + y^2)^2
Ex.3)
