Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.3 Partial Derivatives - 14.3 Exercises: 23

Answer

$$\frac{\partial}{\partial x}f(x,y)=\frac{(ad-bc)y}{(cx+dy)^2};$$ $$\frac{\partial}{\partial y}f(x,y)=\frac{(bc-ad)x}{(cx+dy)^2}.$$

Work Step by Step

Regard $y$ as constant and differentiate with respect to $x$: $$\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x}\left(\frac{ax+by}{cx+dy}\right)=\frac{(ax+by)'_x(cx+dy)-(ax+by)(cx+dy)'_x}{(cx+dy)^2}=\frac{a(cx+dy)-c(ax+by)}{(cx+dy)^2}=\frac{acx+ady-acx-bcy}{(cx+dy)^2}=\frac{(ad-bc)y}{(cx+dy)^2}.$$ Regard $x$ as constant and differentiate with respect to $y$: $$\frac{\partial}{\partial y}f(x,y)=\frac{\partial}{\partial y}\left(\frac{ax+by}{cx+dy}\right)=\frac{(ax+by)'_y(cx+dy)-(ax+by)(cx+dy)'_y}{(cx+dy)^2}=\frac{b(cx+dy)-d(ax+by)}{(cx+dy)^2}=\frac{bcx+bdy-adx-bdy}{(cx+dy)^2}=\frac{(bc-ad)x}{(cx+dy)^2}.$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.