Partial Derivatives of x ln y^2 + y e^z

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Theorem

Let:

$\map f {x, y, z} = x \ln y^2 + y e^z$

Then:

\(\ds \map {f_1} {1, -1, 0}\) \(=\) \(\ds 0\)
\(\ds \map {f_2} {x, x y, y + z}\) \(=\) \(\ds \dfrac 2 y + e^{y + z}\)


Proof

\(\ds \map {f_1} {x, y, z}\) \(=\) \(\ds \ln y^2\) Derivative of Constant Multiple, Derivative of Identity Function, holding $y$ and $z$ constant
\(\ds \leadsto \ \ \) \(\ds \map {f_1} {1, -1, 0}\) \(=\) \(\ds \map \ln {\paren {-1}^2}\) substituting $\tuple {1, -1, 0}$ for $\tuple {x, y, z}$
\(\ds \) \(=\) \(\ds \ln 1\) simplifying
\(\ds \) \(=\) \(\ds 0\) Natural Logarithm of 1 is 0

$\Box$


\(\ds \map {f_2} {x, y, z}\) \(=\) \(\ds \map {\dfrac \partial {\partial y} } {x \ln y^2 + y e^z}\) Definition of Partial Derivative
\(\ds \) \(=\) \(\ds \dfrac x {y^2} \cdot 2 y + e^z\) Derivative of Natural Logarithm, Power Rule for Derivatives, Chain Rule for Derivatives, holding $x$ and $z$ constant
\(\ds \) \(=\) \(\ds \dfrac {2 x} y + e^z\) simplification
\(\ds \leadsto \ \ \) \(\ds \map {f_2} {x, x y, y + z}\) \(=\) \(\ds \dfrac {2 x} {x y} + e^{y + z}\) substituting $\tuple {x, x y, y + z}$ for $\tuple {x, y, z}$
\(\ds \) \(=\) \(\ds \dfrac 2 y + e^{y + z}\) simplifying

$\blacksquare$


Sources