← 2017 Paper 1
UPSC 2017 Maths Optional Paper 1 Q3c — Step-by-Step Solution 15 marks · Section A
Partial derivatives · Calculus · asked 9× in 14 yrs · Read the full method →
Question
If
f ( x , y ) = { x y ( x 2 − y 2 ) x 2 + y 2 , ( x , y ) ≠ ( 0 , 0 ) 0 , ( x , y ) = ( 0 , 0 ) , f(x,y)=\begin{cases}\dfrac{xy(x^2-y^2)}{x^2+y^2}, & (x,y)\ne(0,0)\\ 0, & (x,y)=(0,0),\end{cases} f ( x , y ) = ⎩ ⎨ ⎧ x 2 + y 2 x y ( x 2 − y 2 ) , 0 , ( x , y ) = ( 0 , 0 ) ( x , y ) = ( 0 , 0 ) , calculate ∂ 2 f ∂ x ∂ y \dfrac{\partial^2 f}{\partial x\partial y} ∂ x ∂ y ∂ 2 f ∂ 2 f ∂ y ∂ x \dfrac{\partial^2 f}{\partial y\partial x} ∂ y ∂ x ∂ 2 f ( 0 , 0 ) (0,0) ( 0 , 0 )
Technique
Compute f x ( 0 , y ) f_x(0,y) f x ( 0 , y ) f y ( x , 0 ) f_y(x,0) f y ( x , 0 )
Solution
This is the classic example where f x y ( 0 , 0 ) ≠ f y x ( 0 , 0 ) f_{xy}(0,0)\ne f_{yx}(0,0) f x y ( 0 , 0 ) = f y x ( 0 , 0 ) fails . We must use the difference-quotient definitions of the partials at the origin, not blind differentiation.
Step 1 — First partials on the punctured plane
For ( x , y ) ≠ ( 0 , 0 ) (x,y)\ne(0,0) ( x , y ) = ( 0 , 0 ) f = x y ( x 2 − y 2 ) x 2 + y 2 f=\dfrac{xy(x^2-y^2)}{x^2+y^2} f = x 2 + y 2 x y ( x 2 − y 2 )
f x ( x , y ) = y ( x 4 + 4 x 2 y 2 − y 4 ) ( x 2 + y 2 ) 2 , f y ( x , y ) = x ( x 4 − 4 x 2 y 2 − y 4 ) ( x 2 + y 2 ) 2 . f_x(x,y)=\frac{y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2},\qquad f_y(x,y)=\frac{x(x^4-4x^2y^2-y^4)}{(x^2+y^2)^2}. f x ( x , y ) = ( x 2 + y 2 ) 2 y ( x 4 + 4 x 2 y 2 − y 4 ) , f y ( x , y ) = ( x 2 + y 2 ) 2 x ( x 4 − 4 x 2 y 2 − y 4 ) . Step 2 — Partials along the axes (from the definition)
f ( x , 0 ) = 0 f(x,0)=0 f ( x , 0 ) = 0 f ( 0 , y ) = 0 f(0,y)=0 f ( 0 , y ) = 0 f f f
f x ( 0 , y ) = lim h → 0 f ( h , y ) − f ( 0 , y ) h = lim h → 0 1 h ⋅ h y ( h 2 − y 2 ) h 2 + y 2 = y ( − y 2 ) y 2 = − y ( y ≠ 0 ) , f_x(0,y)=\lim_{h\to0}\frac{f(h,y)-f(0,y)}{h}=\lim_{h\to0}\frac{1}{h}\cdot\frac{hy(h^2-y^2)}{h^2+y^2}=\frac{y(-y^2)}{y^2}=-y\quad(y\ne0), f x ( 0 , y ) = h → 0 lim h f ( h , y ) − f ( 0 , y ) = h → 0 lim h 1 ⋅ h 2 + y 2 h y ( h 2 − y 2 ) = y 2 y ( − y 2 ) = − y ( y = 0 ) , and f x ( 0 , 0 ) = 0 f_x(0,0)=0 f x ( 0 , 0 ) = 0 f x ( 0 , y ) = − y f_x(0,y)=-y f x ( 0 , y ) = − y all y y y
f y ( x , 0 ) = lim k → 0 f ( x , k ) − f ( x , 0 ) k = lim k → 0 1 k ⋅ x k ( x 2 − k 2 ) x 2 + k 2 = x ⋅ x 2 x 2 = x for all x . f_y(x,0)=\lim_{k\to0}\frac{f(x,k)-f(x,0)}{k}=\lim_{k\to0}\frac1k\cdot\frac{xk(x^2-k^2)}{x^2+k^2}=\frac{x\cdot x^2}{x^2}=x\quad\text{for all }x. f y ( x , 0 ) = k → 0 lim k f ( x , k ) − f ( x , 0 ) = k → 0 lim k 1 ⋅ x 2 + k 2 x k ( x 2 − k 2 ) = x 2 x ⋅ x 2 = x for all x . Step 3 — Mixed partials at the origin
∂ 2 f ∂ x ∂ y ( 0 , 0 ) = ∂ ∂ y [ f x ] ( 0 , 0 ) = lim k → 0 f x ( 0 , k ) − f x ( 0 , 0 ) k = lim k → 0 − k − 0 k = − 1. \frac{\partial^2 f}{\partial x\,\partial y}(0,0)=\frac{\partial}{\partial y}\Big[f_x\Big]_{(0,0)} =\lim_{k\to0}\frac{f_x(0,k)-f_x(0,0)}{k}=\lim_{k\to0}\frac{-k-0}{k}=-1. ∂ x ∂ y ∂ 2 f ( 0 , 0 ) = ∂ y ∂ [ f x ] ( 0 , 0 ) = k → 0 lim k f x ( 0 , k ) − f x ( 0 , 0 ) = k → 0 lim k − k − 0 = − 1. ∂ 2 f ∂ y ∂ x ( 0 , 0 ) = ∂ ∂ x [ f y ] ( 0 , 0 ) = lim h → 0 f y ( h , 0 ) − f y ( 0 , 0 ) h = lim h → 0 h − 0 h = + 1. \frac{\partial^2 f}{\partial y\,\partial x}(0,0)=\frac{\partial}{\partial x}\Big[f_y\Big]_{(0,0)} =\lim_{h\to0}\frac{f_y(h,0)-f_y(0,0)}{h}=\lim_{h\to0}\frac{h-0}{h}=+1. ∂ y ∂ x ∂ 2 f ( 0 , 0 ) = ∂ x ∂ [ f y ] ( 0 , 0 ) = h → 0 lim h f y ( h , 0 ) − f y ( 0 , 0 ) = h → 0 lim h h − 0 = + 1. Answer
∂ 2 f ∂ x ∂ y ( 0 , 0 ) = − 1 , ∂ 2 f ∂ y ∂ x ( 0 , 0 ) = + 1. \boxed{\;\frac{\partial^2 f}{\partial x\,\partial y}(0,0)=-1,\qquad \frac{\partial^2 f}{\partial y\,\partial x}(0,0)=+1.\;} ∂ x ∂ y ∂ 2 f ( 0 , 0 ) = − 1 , ∂ y ∂ x ∂ 2 f ( 0 , 0 ) = + 1.