Answer
$$
w_x= y^2z^3 ,\quad w_y= 2xyz^3,\quad w_z=3xy^2z^2.
$$
Work Step by Step
Recall that $(x^n)'=nx^{n-1}$
Since $ w=xy^2z^3$, we have
$$
w_x= y^2z^3 ,\quad w_y= 2xyz^3,\quad w_z=3xy^2z^2.
$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 15 - Differentiation in Several Variables - 15.3 Partial Derivatives - Exercises - Page 781: 37
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