Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.5 - The Chain Rule. - 14.5 Exercise - Page 946: 56


$x^2\dfrac{\partial ^2 f }{\partial x^2} + 2xy \dfrac{\partial ^2 f }{\partial x \partial y} +y^2 \dfrac{\partial ^2 f }{\partial y^2} =n(n-1) f(x,y)$

Work Step by Step

$f_x[\dfrac{d(tx)}{dt}]+f_y[\dfrac{d(ty)}{dt}]=nt^{n-1}f(x,y)$ Here, $\dfrac{d(tx)}{dt}=x$ and $\dfrac{d(ty)}{dt}=y$ and $\dfrac{d^2(tx)}{dt^2}=x$ and $\dfrac{d^2(ty)}{dt^2}=0$ $[x(xf_{xx}+yf_{xy}+(0)f_x]+[y(y(f_{yy}+xf_{xy})+(0)f_y]=n(n-1) t^{n-1} f(x,y)$ This gives: $x^2f_{xx} +xyf_{xy} +xy f_{xy}=n(n-1) t^{n-2} f(x,y)$ Plug in $t=1$ $x^2f_{xx} + 2xyf_{xy} +y^2 f_{yy}=n(n-1) f(x,y)$ Hence, we have: $x^2\dfrac{\partial ^2 f }{\partial x^2} + 2xy \dfrac{\partial ^2 f }{\partial x \partial y} +y^2 \dfrac{\partial ^2 f }{\partial y^2} =n(n-1) f(x,y)$
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