Let $w = \tan ( x y z )$ . Then the differential $d ^ { w }$ is

A) $\sec ^ { 2 } ( x y z ) [ y z d x - x z d y + x y d z ]$
B) $\sec ^ { 2 } ( x y z ) [ y z d x + x z d y + x y d z ]$
C) $\sec ^ { 2 } ( x y z ) [ y z d x + x z d y - x y d z ]$
D) $\sec ^ { 2 } ( x y z ) [ y z d x - x z d y - x y d z ]$
E) $\sec ^ { 2 } ( x y z ) [ - y z d x + x z d y + x y d z ]$

Question

Let $w = \tan ( x y z )$ . Then the differential $d ^ { w }$ is

A) $\sec ^ { 2 } ( x y z ) [ y z d x - x z d y + x y d z ]$
B) $\sec ^ { 2 } ( x y z ) [ y z d x + x z d y + x y d z ]$
C) $\sec ^ { 2 } ( x y z ) [ y z d x + x z d y - x y d z ]$
D) $\sec ^ { 2 } ( x y z ) [ y z d x - x z d y - x y d z ]$
E) $\sec ^ { 2 } ( x y z ) [ - y z d x + x z d y + x y d z ]$

Quizplus · Accepted Answer

To find the differential $dw$, we use the chain rule for differentiation. Given $w = \tan(xyz)$, we first differentiate $w$ with respect to $xyz$, which gives us $\sec^2(xyz)$. Then, we differentiate $xyz$ with respect to $x$, $y$, and $z$ respectively, yielding $yz$, $xz$, and $xy$. Thus, the differential $dw$ is $\sec^2(xyz)[yzdx + xzdy + xydz]$.

Let w=tan⁡(xyz)w = \tan ( x y z )w=tan(xyz) Then the Differential

Let $w = \tan ( x y z )$ Then the Differential