Question

for \( f_{z} \), we have \[ \begin{aligned} f_{y}(x, y, z) & =f_{z}\left(\frac{y z^{2}}{x^{3}}\right)+f_{z}\left[\tan \left(2 x^{2}+4 y-z\right)\right] \\ & =f_{z}\left(y^{2} x^{-3}\right)+f_{z}\left[\tan \left(2 x^{2}+4 y-z\right)\right] \\ & =2 y z x^{-3}+\sec ^{2}\left(2 x^{2}+4 y-z\right) \cdot(-1) \\ & =\frac{2 y z}{x^{3}}-\sec ^{2}\left(2 x^{2}+4 y-z\right) \end{aligned} \]

          for \( f_{z} \), we have
\[
\begin{aligned}
f_{y}(x, y, z) & =f_{z}\left(\frac{y z^{2}}{x^{3}}\right)+f_{z}\left[\tan \left(2 x^{2}+4 y-z\right)\right] \\
& =f_{z}\left(y^{2} x^{-3}\right)+f_{z}\left[\tan \left(2 x^{2}+4 y-z\right)\right] \\
& =2 y z x^{-3}+\sec ^{2}\left(2 x^{2}+4 y-z\right) \cdot(-1) \\
& =\frac{2 y z}{x^{3}}-\sec ^{2}\left(2 x^{2}+4 y-z\right)
\end{aligned}
\]

for fz, we have

fy(x, y, z) =fz((y z^2)/(x^3))+fz[tan(2 x^2+4 y-z)]
=fz(y^2 x^-3)+fz[tan(2 x^2+4 y-z)]
=2 y z x^-3+sec ^2(2 x^2+4 y-z) ·(-1)
=(2 y z)/(x^3)-sec ^2(2 x^2+4 y-z)

Added by Catherine M.

Calculus

Earl W. Swokowski 5th Edition

Chapter 16

Instant Answer

Solved by Expert Ronald Prasad

Step 1

The function is given as: \[ f_y(x, y, z) = f_z\left(\frac{y z^2}{x^3}\right) + f_z\left[\tan(2x^2 + 4y - z)\right] \] Show more…

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Thanks for your feedback!

for \( f_{z} \), we have \[ \begin{aligned} f_{y}(x, y, z) & =f_{z}\left(\frac{y z^{2}}{x^{3}}\right)+f_{z}\left[\tan \left(2 x^{2}+4 y-z\right)\right] \\ & =f_{z}\left(y^{2} x^{-3}\right)+f_{z}\left[\tan \left(2 x^{2}+4 y-z\right)\right] \\ & =2 y z x^{-3}+\sec ^{2}\left(2 x^{2}+4 y-z\right) \cdot(-1) \\ & =\frac{2 y z}{x^{3}}-\sec ^{2}\left(2 x^{2}+4 y-z\right) \end{aligned} \]